//セグ木(二分探索)O(N(logN)^2)

#include <iostream>
#include <algorithm>
#include <atcoder/all>
using namespace std;
using namespace atcoder;
using ll = long long;

int op(int a, int b) { return max(a, b); }
int e() { return -1; }

int main(){
  int N, Q;
  cin >> N >> Q;
  vector<int> A(N, 0);
  for(int i = 0; i < N; i++) cin >> A[i];
  segtree<int, op, e> seg(A);
  for(int q = 0; q < Q; q++){
    int c, target;
    cin >> c >> target;
    target++;
    if(c == 1){
      int l = 0, r = N;
      while(l < r){
        int m = (l + r + 1) / 2;
        auto ret = seg.prod(l, m);
        if(ret >= target) r = m - 1;
        else l = m;
      }
      if(l >= N){
        cout << -1 << endl;
      }else{
        cout << l + 1 << endl;
        seg.set(l, 0);
      }
    }
    if(c == 2){
      int l = 0, r = N;
      while(l < r){
        int m = (l + r) / 2;
        auto ret = seg.prod(m, r);
        if(ret >= target) l = m + 1;
        else r = m;
      }
      l--;
      if(l < 0){
        cout << -1 << endl;
      }else{
        cout << l + 1 << endl;
        seg.set(l, 0);
      }
    }
  }
  return 0;
}