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 INF = 1e9;
const long long LINF = 1e18;
const long double EPS = 1e-10;
const long double PI = 3.141592653589;
const int N = 1e6;
int n, m;
int a[N];
int nxt[N];
int main() {
  ios::sync_with_stdio(false);
  float tbegin = clock();
  srand(42);
  if ("" != "") {
    freopen(
        ""
        ".in",
        "r", stdin);
    freopen(
        ""
        ".out",
        "w", stdout);
  }
  scanf("%d%d", &n, &m);
  for (int i = 0; i < n; i++) {
    scanf("%d", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    int j = i;
    while (j + 1 < n && a[j + 1] == a[i]) {
      j++;
    }
    for (int k = i; k <= j; k++) {
      nxt[k] = j + 1;
    }
    i = j;
  }
  for (int i = 0; i < m; i++) {
    int l, r, x;
    scanf("%d%d%d", &l, &r, &x);
    l--;
    r--;
    int ans = -1;
    if (a[l] != x) {
      ans = l + 1;
    } else if (nxt[l] <= r) {
      ans = nxt[l] + 1;
    }
    printf("%d\n", ans);
  }
  return 0;
}