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 SZ = 400;
int N, Q;
long long A[500000];
struct bucket {
  int l, r;
  long long lazy;
  vector<pair<long long, int>> v;
  bucket() {
    l = 0x3f3f3f3f;
    r = -0x3f3f3f3f;
  }
  void down() {
    v.clear();
    for (int i = l; i <= r; i++) {
      A[i] += lazy;
      v.push_back(make_pair(A[i], i));
    }
    sort(v.begin(), v.end());
    lazy = 0;
  }
  int find_first(long long x) {
    x -= lazy;
    auto it = lower_bound(v.begin(), v.end(), make_pair(x, 0));
    if (it == v.end() || it->first != x) return -1;
    return it->second;
  }
  int find_last(long long x) {
    x -= lazy;
    auto it = lower_bound(v.begin(), v.end(), make_pair(x + 1, 0));
    if (it == v.begin()) return -1;
    --it;
    if (it->first != x) return -1;
    return it->second;
  }
  void update(int p, int q, long long v) {
    if (p <= l && r <= q) {
      lazy += v;
      return;
    }
    if (lazy) down();
    for (int i = max(p, l); i <= min(q, r); i++) A[i] += v;
    down();
  }
} buckets[500000 / SZ + 5];
int main() {
  scanf("%d%d", &N, &Q);
  for (int i = 0; i < N; i++) {
    scanf("%lld", A + i);
    int idx = i / SZ;
    buckets[idx].l = min(buckets[idx].l, i);
    buckets[idx].r = max(buckets[idx].r, i);
  }
  for (int i = 0; i < N; i++)
    if (i == 0 || i / SZ != (i - 1) / SZ) buckets[i / SZ].down();
  int a, b, c, op;
  for (int i = 0; i < Q; i++) {
    scanf("%d", &op);
    if (op == 1) {
      scanf("%d%d%d", &a, &b, &c);
      a--, b--;
      for (int i = a / SZ; i <= b / SZ; i++) buckets[i].update(a, b, c);
    } else {
      scanf("%d", &a);
      int first = -1;
      for (int i = 0; first == -1 && i <= (N - 1) / SZ; i++)
        first = buckets[i].find_first(a);
      if (first == -1)
        printf("-1\n");
      else {
        int last = -1;
        for (int i = (N - 1) / SZ; last == -1 && i >= 0; i--)
          last = buckets[i].find_last(a);
        printf("%d\n", last - first);
      }
    }
  }
  return 0;
}