//幽默题
/*
思路1：
转平面直角坐标系，贯穿于同一根45度斜线的点相互连边

思路2：
拆绝对值
红边：若干个团
蓝边：若干个团
不存在重合的红边和蓝边
每个点有红色数字和蓝色数字
不存在两个红色数字和蓝数都相同的点
有若干个组，每组中选择奇数/偶数个点，让它们可以两两匹配

树链剖分我们喜欢你
*/
#include <bits/stdc++.h>//幽默题
using namespace std;
typedef long long ll;
#define rep(i, a, b) for(ll i=a;i<=b;i++)
#define rrep(i, a, b) for(ll i=a;i>=b;i--)
const ll N=5e5+9;
ll T, n, a[N];
struct point{
    ll red, blue, id;
    bool operator<(const point &b)const{
        return red<b.red||red==b.red&&blue<b.blue;
    }
}p[N];
ll bred[N], bblue[N],nR, nB;
bool vst[N], vot[N];
ll cnt[N];//cnt[i]表示红色为i的点有多少个
vector <ll> roast[N], tog[N], tot[N];
vector <pair<ll, ll> > pog[N];//po表示一条边对应的是那两个点之间的抵消
pair <ll, ll> po[N];//po[i]表示i和它父亲的这一条边
void dfs_findtree(ll u){
    vst[u]=1;
    for(ll i=0;i<tog[u].size();i++){ll v=tog[u][i];
        if(vst[v])continue;
        tot[u].emplace_back(v);
        tot[v].emplace_back(u);
        po[v].first=pog[u][i].first;
        po[v].second=pog[u][i].second;

        dfs_findtree(v);
    }    
}
ll need[N];
ll inblue[N];
bool dfs_solve(ll u, ll fa){//返回又没有剩余
    // cout << 'u' << u << '\n';
    vot[u]=1;
    ll lastfail=0;
    for(ll v:tot[u]){
        if(v==fa)continue;
        bool x=dfs_solve(v, u);
        if(x){
            if(lastfail==0)lastfail=v;
            else {need[v]^=1;need[lastfail]^=1;lastfail=0;}
        }
    }
    if(cnt[u]&1){
        if(lastfail==0)return 1;
        else {need[lastfail]^=1;return 0;}
    }else{
        if(lastfail==0)return 0;
        else {need[lastfail]^=1;return 1;}
    }
}
void clear(){
    fill(need, need+n+5, 0);
    fill(inblue, inblue+n+5, 0);
    fill(bred, bred+n+5, 0);
    fill(bblue, bblue+n+5, 0);
    fill(vst, vst+n+5, 0);
    fill(vot, vot+n+5, 0);
    nR=0;nB=0;
    fill(cnt, cnt+n+5, 0);
    rep(i, 1, n+3){
        roast[i].clear();
        tog[i].clear();

        tot[i].clear();

        pog[i].clear();

        po[i]=make_pair(0, 0);

    }
}
int main(){
 //   freopen("matching.in", "r", stdin);
  //  freopen("matching.out", "w", stdout);
    ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
    cin >> T;
    while(T--){
        cin >> n;
        rep(i, 1, n){
            cin >> a[i];
            p[i].red=i-a[i];
            p[i].blue=i+a[i];
            p[i].id=i;
            bred[i]=p[i].red;
            bblue[i]=p[i].blue;
        }
        sort(bred+1, bred+n+1);
        nR=unique(bred+1, bred+n+1)-(bred+1);
        sort(bblue+1, bblue+n+1);
        nB=unique(bblue+1, bblue+n+1)-(bblue+1);
        rep(i, 1, n){
            p[i].red=lower_bound(bred+1, bred+nR+1, p[i].red)-bred;
            p[i].blue=lower_bound(bblue+1, bblue+nB+1, p[i].blue)-bblue;
            // cout << p[i].red << ' ' << p[i].blue << '\n';
        }
        sort(p+1, p+n+1);
        rep(i, 1, n){
            cnt[p[i].red]++;
            roast[p[i].blue].emplace_back(i);
            // cout << i << "is in " << p[i].blue << endl;
        }
        // rep(i, 1, nR){
        //     cout << i << ' ' << cnt[i] << endl;
        // }
        rep(i, 1, nB){
            for(ll j=1;j<roast[i].size();j++){
                // cout << "in" << i << ' ' << roast[i][j] << '\n';
                tog[p[roast[i][j-1]].red].emplace_back(p[roast[i][j]].red);
                tog[p[roast[i][j]].red].emplace_back(p[roast[i][j-1]].red);
                pog[p[roast[i][j-1]].red].emplace_back(make_pair(roast[i][j-1], roast[i][j]));
                pog[p[roast[i][j]].red].emplace_back(make_pair(roast[i][j-1], roast[i][j]));
            }
        }
        rep(i, 1, nR){
            if(!vst[i]){
                dfs_findtree(i);
            }
        }
        bool ok=1;
        rep(i, 1, nR){
            if(!vot[i])
            if(dfs_solve(i, 0)){
                ok=0;break;
            }
            // cout << '\n';
        }
        if(ok==0){
            cout << "No" << '\n';
        }else{
            cout << "Yes" << '\n';
            rep(i, 1, nR){
                // cout << need[i] << '/';
                if(need[i]){
                    inblue[po[i].first]^=1;
                    inblue[po[i].second]^=1;
                }
            }
            // cout << '\n';
            // rep(i, 1, n){
            //     cout << inblue[i] << '|';
            // }
            // cout << '\n';
            rep(i, 1, nB){
                ll lastib=0;
                for(ll j=0;j<roast[i].size();j++){
                    ll u=roast[i][j];
                    if(inblue[u]){
                        if(lastib==0)lastib=u;
                        else {cout << p[lastib].id << ' ' << p[u].id << '\n';lastib=0;}
                    }
                }
            }
            ll lastd=0;
            rep(i, 1, n){
                if(p[i].red!=p[i-1].red){
                    lastd=0;
                }
                if(!inblue[i]){
                    if(lastd==0)lastd=i;
                    else {cout << p[lastd].id << ' ' << p[i].id << '\n';lastd=0;}
                }
            }
        }
        clear();
    }
    return 0;
}