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;
const int maxn = 2e5 + 5;
int low[maxn];
int num[maxn];
int n, m;
struct cx {
  int L;
  int R;
  int V;
} cx1[maxn];
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) {
    scanf("%d", &num[i]);
    if (i != 1) {
      if (num[i] == num[i - 1])
        low[i] = low[i - 1];
      else
        low[i] = i - 1;
    } else
      low[i] = 0;
  }
  for (int i = 0; i < m; i++) {
    scanf("%d%d%d", &cx1[i].L, &cx1[i].R, &cx1[i].V);
  }
  for (int i = 0; i < m; i++) {
    if (num[cx1[i].R] != cx1[i].V)
      printf("%d\n", cx1[i].R);
    else {
      if (low[cx1[i].R] < cx1[i].L)
        printf("-1\n");
      else
        printf("%d\n", low[cx1[i].R]);
    }
  }
  return 0;
}