Problem: You are given array a with n integers and m queries. The i-th query is given with three integers li, ri, xi.

For the i-th query find any position pi (li ≤ pi ≤ ri) so that api ≠ xi.

Input

The first line contains two integers n, m (1 ≤ n, m ≤ 2·105) — the number of elements in a and the number of queries.

The second line contains n integers ai (1 ≤ ai ≤ 106) — the elements of the array a.

Each of the next m lines contains three integers li, ri, xi (1 ≤ li ≤ ri ≤ n, 1 ≤ xi ≤ 106) — the parameters of the i-th query.

Output

Print m lines. On the i-th line print integer pi — the position of any number not equal to xi in segment [li, ri] or the value  - 1 if there is no such number.

Examples

Input

6 4
1 2 1 1 3 5
1 4 1
2 6 2
3 4 1
3 4 2


Output

2
6
-1
4

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
const int Nmax = 100005;
const int Nmic = 105;
const int N = 1005;
using namespace std;
int st, dr, R, L, l, r, x, n, q, i, mij, p;
vector<int> v[10 * Nmax];
int main() {
  scanf("%d%d", &n, &q);
  for (i = 1; i <= n; ++i) {
    scanf("%d", &x);
    v[x].push_back(i);
  }
  for (i = 0; i <= Nmax * 10 - 4; ++i) sort(v[i].begin(), v[i].end());
  for (i = 1; i <= q; ++i) {
    scanf("%d%d%d", &l, &r, &x);
    if (v[x].size() == 0) {
      printf("%d\n", l);
      continue;
    }
    st = 1;
    dr = v[x].size();
    while (st <= dr) {
      mij = (st + dr) / 2;
      if (v[x][mij - 1] < l)
        st = mij + 1;
      else
        dr = mij - 1;
    }
    L = st;
    st = 1;
    dr = v[x].size();
    while (st <= dr) {
      mij = (st + dr) / 2;
      if (v[x][mij - 1] <= r)
        st = mij + 1;
      else
        dr = mij - 1;
    }
    R = dr;
    if (!R || L == v[x].size() + 1) {
      printf("%d\n", l);
      continue;
    }
    if (L - R + 1 == l - r + 1) {
      printf("-1\n");
      continue;
    }
    if (v[x][L - 1] != l) {
      printf("%d\n", l);
      continue;
    }
    st = L;
    dr = R;
    p = l - L;
    while (st <= dr) {
      mij = (st + dr) / 2;
      if (v[x][mij - 1] != p + mij)
        dr = mij - 1;
      else
        st = mij + 1;
    }
    printf("%d\n", v[x][dr - 1] + 1);
  }
  return 0;
}