#include<bits/stdc++.h>
#define ll long long int
#define db double
#define tp ll
#define ptp pair<tp,tp>
#define pdb pair<db,db>
#define umap unordered_map
#define ed '\n'


using namespace std;

tp n, num, sum, flag, ans, tmp, ls, len, wz, pos, m, k, p, q;
tp sqrtn;
tp lt, rt, mid;
tp min1, min2, min3, max1, max2, max3;
string s;
tp h, w, t;

namespace anymod_poly {
    tp mod;

    tp ksm(tp a, tp b, tp p) {
        tp lsans = 1;
        while (b) {
            if (b & 1ll) {
                lsans = (lsans * a) % p;
            }
            a = (a * a) % p;
            b >>= 1;
        }
        return lsans;
    }

    tp inv(tp x, tp mod) { return ksm(x, mod - 2, mod); }

    const tp mod1 = 998244353, mod2 = 1004535809, mod3 = 469762049, G = 3;
    const ll mod_1_2 = static_cast<ll> (mod1) * mod2;
    const tp inv_1 = inv(mod1, mod2), inv_2 = inv(mod_1_2 % mod3, mod3);
    struct Int {
        ll A, B, C;
        explicit Int() { }
        explicit Int(ll __num) : A(__num), B(__num), C(__num) { }
        explicit Int(ll __A, ll __B, ll __C) : A(__A), B(__B), C(__C) { }
        static Int reduce(const Int& x) {
            return Int(x.A + (x.A >> 31 & mod1), x.B + (x.B >> 31 & mod2), x.C + (x.C >> 31 & mod3));
        }
        friend Int operator + (const Int& lhs, const Int& rhs) {
            return reduce(Int(lhs.A + rhs.A - mod1, lhs.B + rhs.B - mod2, lhs.C + rhs.C - mod3));
        }
        friend Int operator - (const Int& lhs, const Int& rhs) {
            return reduce(Int(lhs.A - rhs.A, lhs.B - rhs.B, lhs.C - rhs.C));
        }
        friend Int operator * (const Int& lhs, const Int& rhs) {
            return Int(static_cast<ll> (lhs.A) * rhs.A % mod1, static_cast<ll> (lhs.B) * rhs.B % mod2, static_cast<ll> (lhs.C) * rhs.C % mod3);
        }
        tp get() {
            ll x = static_cast<ll> (B - A + mod2) % mod2 * inv_1 % mod2 * mod1 + A;
            return (static_cast<ll> (C - x % mod3 + mod3) % mod3 * inv_2 % mod3 * (mod_1_2 % mod) % mod + x) % mod;
        }
    };

    tp lim, s;
    vector<tp>rev;
    vector<Int>Wn;
    void poly_init(tp n) {
        s = -1, lim = 1;
        while (lim < n) lim <<= 1, ++s;
        rev = vector<tp>(lim + 5);
        Wn = vector<Int>(lim + 5);
        for (register tp i = 1; i < lim; ++i) rev[i] = rev[i >> 1] >> 1 | (i & 1) << s;
        const Int t(ksm(G, (mod1 - 1) / lim, mod1), ksm(G, (mod2 - 1) / lim, mod2), ksm(G, (mod3 - 1) / lim, mod3));
        Wn[0] = Int(1);
        for (auto it = Wn.begin(); it != Wn.begin() + lim; ++it) *(it + 1) = *it * t;
    }
    void ntt(Int* A, const tp op) {//传入的数组尺寸至少要比lim大
        for (register tp i = 1; i < lim; ++i) if (i < rev[i]) std::swap(A[i], A[rev[i]]);
        for (register tp mid = 1; mid < lim; mid <<= 1) {
            const tp t = lim / mid >> 1;
            for (register tp i = 0; i < lim; i += mid << 1) {
                for (register tp j = 0; j < mid; ++j) {
                    const Int W = op ? Wn[t * j] : Wn[lim - t * j];
                    const Int X = A[i + j], Y = A[i + j + mid] * W;
                    A[i + j] = X + Y, A[i + j + mid] = X - Y;
                }
            }
        }
        if (!op) {
            const Int ilim(inv(lim, mod1), inv(lim, mod2), inv(lim, mod3));
            for (register Int* i = A; i != A + lim; ++i) *i = (*i) * ilim;
        }
    }

    void mul(Int* a, Int* b, Int* c) {
        for (tp i = 0; i < lim; ++i) c[i] = a[i] * b[i];
        ntt(c, 0);
        for (tp i = m + 1; i < lim; i++) {//去除大于m的阶
            c[i] = Int(0);
        }
        ntt(c, 1);
    }

}

using namespace anymod_poly;

Int b[520], bb[520];
Int a[1505][520];
Int pre[42][1505][520];
ptp qr[600005];
Int qk[42][42][42][520];


void init() {

}

void clear() {

}

void solve() {
    mod = 1e9 + 7;
    cin >> n >> m >> q;
    sqrtn = sqrt(n);
    poly_init(2 * m + 1);

    for (tp i = 0; i <= m; i++) {
        cin >> ls;
        b[i] = Int(ls % mod);
        bb[m - i] = b[i];
    }
    //将bb转化
    ntt(bb, 1);
    for (tp i = 1; i <= n; i++) {
        for (tp j = 0; j <= m; j++) {
            cin >> ls;
            a[i][j] = Int(ls % mod);
        }
    }

    for (tp i = 1; i <= n; i++) {
        tp id = (i - 1) / sqrtn + 1;
        for (tp j = 0; j <= m; j++) {
            qk[id][i - (id - 1) * sqrtn][i - (id - 1) * sqrtn][j] = a[i][j];
        }
        ntt(qk[id][i - (id - 1) * sqrtn][i - (id - 1) * sqrtn], 1);
    }
    //将a[i]转化
    for (tp i = 1; i <= n; i++) {
        ntt(a[i], 1);
        
    }

    for (tp i = 1; i <= q; i++) {
        cin >> qr[i].first >> qr[i].second;
    }


    for (tp i = 1; (i-1) * sqrtn <= n; i++) {
        pre[i][(i - 1) * sqrtn][0] = Int(1);
        ntt(pre[i][(i - 1) * sqrtn], 1);
    }
    for (tp i = 1; i <= n; i++) {
        tp id = (i - 1) / sqrtn + 1;
        for (tp j = 1; j <= id; j++) {
            mul(pre[j][i - 1], a[i], pre[j][i]);
        }
    }

    for (tp id = 1; id * sqrtn <= n; id++) {
        for (tp len = 2; len <= sqrtn; len++) {
            for (tp i = 1; i + len - 1 <= sqrtn; i++) {
                mul(qk[id][i][i], qk[id][i + 1][i + len - 1], qk[id][i][i + len - 1]);
            }
        }
    }
    for (tp id = 1; id * sqrtn <= n; id++) {
        for (tp len = 1; len <= sqrtn; len++) {
            for (tp i = 1; i + len - 1 <= sqrtn; i++) {
                mul(qk[id][i][i+len-1],bb,qk[id][i][i+len-1]);
            }
        }
    }

    for (tp id = 1; id * sqrtn <= n; id++) {
        for (tp len = 1; len <= sqrtn; len++) {
            for (tp i = 1; i + len - 1 <= sqrtn; i++) {
                ntt(qk[id][i][i + len - 1], 0);
            }
        }
    }

    for (tp i = 1; i <= n; i++) {
        tp id = (i - 1) / sqrtn + 1;
        for (tp j = 1; j <= id; j++) {
            ntt(pre[j][i], 0);
        }
    }
    

    for (tp i = 1; i <= q; i++) {
        lt = qr[i].first; rt = qr[i].second;
        tp lid = (lt - 1) / sqrtn + 1;
        tp rid = (rt - 1) / sqrtn + 1;
        if (lid == rid) {
            cout << qk[lid][lt - (lid - 1) * sqrtn][rt - (rid - 1) * sqrtn][m].get() << ed;
        }
        else {
            ans = 0;
            for (tp j = 0; j <= m; j++) {
                ans = (ans + qk[lid][lt - (lid - 1) * sqrtn][sqrtn][j].get() * pre[lid + 1][rt][m - j].get()) % mod;
            }
            cout << ans << ed;
        }
    }

}

int main(void) {
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int _t = 1;
    //cin >> _t;
    init();
    while (_t--) {
        solve();
    }
}
//样例
/*

3 3 3
4 3 2 1
0 1 0 1000000007
0 500000004 0 500000004
0 0 500000004 500000004
1 1
1 2
1 3


3 3 6
4 3 2 1
1000000007 0 1 0
1000000007 1 0 0
1000000007 0 1 0
1 1
1 2
1 3
2 2
2 3
3 3


*/