Problem: Professor GukiZ was playing with arrays again and accidentally discovered new function, which he called GukiZiana. For given array a, indexed with integers from 1 to n, and number y, GukiZiana(a, y) represents maximum value of j - i, such that aj = ai = y. If there is no y as an element in a, then GukiZiana(a, y) is equal to  - 1. GukiZ also prepared a problem for you. This time, you have two types of queries: 

  1. First type has form 1 l r x and asks you to increase values of all ai such that l ≤ i ≤ r by the non-negative integer x. 
  2. Second type has form 2 y and asks you to find value of GukiZiana(a, y). 



For each query of type 2, print the answer and make GukiZ happy!

Input

The first line contains two integers n, q (1 ≤ n ≤ 5 * 105, 1 ≤ q ≤ 5 * 104), size of array a, and the number of queries. 

The second line contains n integers a1, a2, ... an (1 ≤ ai ≤ 109), forming an array a. 

Each of next q lines contain either four or two numbers, as described in statement:

If line starts with 1, then the query looks like 1 l r x (1 ≤ l ≤ r ≤ n, 0 ≤ x ≤ 109), first type query.

If line starts with 2, then th query looks like 2 y (1 ≤ y ≤ 109), second type query.

Output

For each query of type 2, print the value of GukiZiana(a, y), for y value for that query.

Examples

Input

4 3
1 2 3 4
1 1 2 1
1 1 1 1
2 3


Output

2


Input

2 3
1 2
1 2 2 1
2 3
2 4


Output

0
-1

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 500005;
const int maxb = 715;
int n, q;
long long a[maxn], lzy[maxb];
int pl[maxb], pr[maxb], bcnt, bsiz, bid[maxn];
multiset<long long> S[maxb];
inline int read() {
  int res = 0, f_f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f_f = -1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9')
    res = (res << 3) + (res << 1) + (ch - '0'), ch = getchar();
  return res * f_f;
}
inline void update(int x, int y, int v) {
  if (bid[x] != bid[y]) return;
  int id = bid[x];
  if (x > y) swap(x, y);
  for (int i = x; i <= y; i++) {
    S[id].erase(S[id].find(a[i]));
    a[i] += v;
    S[id].insert(a[i]);
  }
}
inline void modify(int x, int y, int v) {
  int l = bid[x], r = bid[y];
  if (l == r) {
    update(x, y, v);
    return;
  }
  update(x, pr[l], v), update(pl[r], y, v);
  for (int i = l + 1; i < r; i++) {
    lzy[i] += v;
  }
}
inline int query(long long x) {
  int l = -1, r = -1;
  for (int i = 1; i <= bcnt; i++) {
    long long k = x - lzy[i];
    if (S[i].find(k) == S[i].end()) continue;
    r = i;
    if (l == -1) l = i;
  }
  if (l == -1) return -1;
  for (int i = pl[l]; i <= pr[l]; i++) {
    if (a[i] + lzy[l] != x) continue;
    l = i;
    break;
  }
  for (int i = pr[r]; i >= pl[r]; i--) {
    if (a[i] + lzy[r] != x) continue;
    r = i;
    break;
  }
  return r - l;
}
int main() {
  n = read(), q = read(), bsiz = (int)sqrt(n) + 1;
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1, j = 1; j <= n; i++, j++) {
    pl[i] = j, j += bsiz - 1, pr[i] = min(j, n), bcnt = i;
    for (int k = pl[i]; k <= pr[i]; k++) {
      bid[k] = i, S[i].insert(a[k]);
    }
  }
  while (q--) {
    int opt = read();
    if (opt == 1) {
      int x = read(), y = read(), v = read();
      modify(x, y, v);
    } else {
      int x = read();
      printf("%d\n", query(x));
    }
  }
  return 0;
}