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 N = 2e5 + 3;
int n, m, arr[N], a, b, x, mn[4 * N], mx[4 * N], q;
void build1(int p, int l, int r) {
  if (l == r) {
    mn[p] = l;
    return;
  }
  int mid = (l + r) / 2;
  build1(2 * p, l, mid);
  build1(2 * p + 1, mid + 1, r);
  if (arr[mn[2 * p]] <= arr[mn[2 * p + 1]])
    mn[p] = mn[2 * p];
  else
    mn[p] = mn[2 * p + 1];
}
void build2(int p, int l, int r) {
  if (l == r) {
    mx[p] = l;
    return;
  }
  int mid = (l + r) / 2;
  build2(2 * p, l, mid);
  build2(2 * p + 1, mid + 1, r);
  if (arr[mx[2 * p]] >= arr[mx[2 * p + 1]])
    mx[p] = mx[2 * p];
  else
    mx[p] = mx[2 * p + 1];
}
int ans1(int p, int l, int r) {
  if (a <= l && r <= b) return mn[p];
  if (l > b || r < a) return -1;
  int mid = (l + r) / 2;
  int il = ans1(2 * p, l, mid);
  int ir = ans1(2 * p + 1, mid + 1, r);
  if (il == -1) return ir;
  if (ir == -1) return il;
  if (arr[il] <= arr[ir])
    return il;
  else
    return ir;
}
int ans2(int p, int l, int r) {
  if (a <= l && r <= b) {
    return mx[p];
  }
  if (l > b || r < a) return -1;
  int mid = (l + r) / 2;
  int il = ans2(2 * p, l, mid);
  int ir = ans2(2 * p + 1, mid + 1, r);
  if (il == -1) return ir;
  if (ir == -1) return il;
  if (arr[il] >= arr[ir])
    return il;
  else
    return ir;
}
int main() {
  scanf("%d%d", &n, &q);
  for (int i = 0; i < n; i++) scanf("%d", arr + i);
  build1(1, 0, n - 1);
  build2(1, 0, n - 1);
  while (q--) {
    scanf("%d%d%d", &a, &b, &x);
    --a;
    --b;
    int u = ans1(1, 0, n - 1);
    int v = ans2(1, 0, n - 1);
    if (arr[u] == arr[v] && arr[u] == x) {
      printf("-1\n");
      continue;
    }
    if (arr[u] != x) {
      printf("%d\n", u + 1);
    } else
      printf("%d\n", v + 1);
  }
  return 0;
}