#pragma GCC optimize(2)
#pragma GCC optimize(3,"Ofast","inline")
#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL
#include "debug.h"
#else
#define debug(...) 42;
#define endl '\n'
#endif

#define ll long long
// #define int long long

const int Maxl = 0, Maxr = 1e18 + 10;
struct d_tree {
    struct op {
        ll num;
        int ls, rs;
    } tre[40004000];
    int cnt = 0;

    //使用前新建一个根点'1',即使用一下build()
    //
    int build()//新建节点
    {   ++cnt;
        tre[cnt].num = tre[cnt].ls = tre[cnt].rs = 0;
        return cnt;
    }

    int upd(int x, ll pos, int val, d_tree &tr, ll l = Maxl, ll r = Maxr)
    {   if (!x)x = tr.build();
        if (l == r) {tre[x].num += val; return x;}
        ll mid = (l + r) >> 1;
        if (pos <= mid)tre[x].ls = tr.upd(tre[x].ls, pos, val, tr, l, mid);
        if (pos > mid)tre[x].rs = tr.upd(tre[x].rs, pos, val, tr, mid + 1, r);
        tre[x].num = tre[tre[x].ls].num + tre[tre[x].rs].num;
        return x;
    }
    ll ask(int x, ll ql, ll qr, d_tree &tr, ll l = Maxl, ll r = Maxr)
    {   if (!x)return 0;
        if (l >= ql && r <= qr) {
            return tre[x].num;
        }
        ll mid = (l + r) >> 1, ans = 0;
        if (ql <= mid)ans += tr.ask(tre[x].ls, ql, qr, tr, l, mid);
        if (qr > mid)ans += tr.ask(tre[x].rs, ql, qr, tr, mid + 1, r);
        return ans;
    }
} tr;
ll nex(__int128 x, ll n) {
    __int128 ans = 0;
    for (int i = 0; i <= 62; ++i) {
        if (((x >> i) & 1) && (((x >> (i + 1)) & 1) == 0)) {
            ans += 1ll << (i + 1);
            for (int j = i + 2; j <= 62; ++j) {
                if ((x >> j) & 1) {
                    ans += 1ll << j;
                }
            }
            break;
        }
    }
    if (ans > n)return 0;
    return ans;
}
signed main()
{
    ios::sync_with_stdio(false); cin.tie(nullptr); cout.tie(nullptr);
    int m, val, op;
    ll n, l, r, x;
    cin >> n >> m;

    tr.build();

    for (int i = 1; i <= m; ++i) {
        cin >> op;
        if (op == 1) {
            cin >> x >> val;
            while (x <= n && x) {
                tr.upd(1, x, val, tr);
                x = nex(x, n);
            }
        }
        else {
            cin >> l >> r;
            cout << tr.ask(1, l, r, tr) << "\n";
        }
    }
}