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>
using namespace std;
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  vector<int> a(n);
  for (int &x : a) {
    scanf("%d", &x);
  }
  vector<int> b(n);
  b[0] = 1;
  for (int i = 1; i < n; ++i) {
    if (a[i] == a[i - 1]) {
      b[i] = b[i - 1];
    } else {
      b[i] = 1 - b[i - 1];
    }
  }
  vector<int> next0(n);
  vector<int> next1(n);
  next0[n - 1] = (b[n - 1] == 1 ? -1 : n - 1);
  for (int i = n - 2; i >= 0; --i) {
    if (b[i] == 1) {
      next0[i] = next0[i + 1];
    } else {
      next0[i] = i;
    }
  }
  next1[n - 1] = (b[n - 1] == 0 ? -1 : n - 1);
  for (int i = n - 2; i >= 0; --i) {
    if (b[i] == 0) {
      next1[i] = next1[i + 1];
    } else {
      next1[i] = i;
    }
  }
  while (m--) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    --l;
    --r;
    if (b[l] == 0) {
      if (next1[l] > r || next1[l] == -1) {
        if (x != a[l]) {
          printf("%d\n", l + 1);
        } else {
          printf("-1\n");
        }
      } else {
        if (a[next1[l]] == x) {
          printf("%d\n", l + 1);
        } else {
          printf("%d\n", next1[l] + 1);
        }
      }
    } else {
      if (next0[l] > r || next0[l] == -1) {
        if (x != a[l]) {
          printf("%d\n", l + 1);
        } else {
          printf("-1\n");
        }
      } else {
        if (a[next0[l]] == x) {
          printf("%d\n", l + 1);
        } else {
          printf("%d\n", next0[l] + 1);
        }
      }
    }
  }
  return 0;
}