#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int maxn = 200010;
struct node {
  int val[5][5];
  node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        val[i][j] = INF;
      }
    }
  }
  node operator+(node x) {
    node ans;
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        for (int k = 0; k < 5; ++k) {
          ans.val[i][j] = min(ans.val[i][j], val[i][k] + x.val[k][j]);
        }
      }
    }
    return ans;
  }
} tree[maxn << 2];
int n, q;
char num[maxn];
char ch[] = {'2', '0', '1', '7', '6'};
int Find(char c) {
  for (int i = 0; i < 5; i++) {
    if (ch[i] == c) return i;
  }
  return -1;
}
node push_up(node &a, node &b) {
  node res;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        res.val[i][j] = min(res.val[i][j], a.val[i][k] + b.val[k][j]);
      }
    }
  }
  return res;
}
void build_tree(int rt, int l, int r) {
  if (l == r) {
    int f = Find(num[l]);
    for (int i = 0; i < 5; i++) {
      tree[rt].val[i][i] = 0;
    }
    if (f != -1 && f < 4) {
      tree[rt].val[f][f + 1] = 0;
      tree[rt].val[f][f] = 1;
    } else if (f == 4) {
      tree[rt].val[3][3] = tree[rt].val[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build_tree(rt << 1, l, mid);
  build_tree(rt << 1 | 1, mid + 1, r);
  tree[rt] = push_up(tree[rt << 1], tree[rt << 1 | 1]);
}
node query(int rt, int L, int R, int l, int r) {
  if (l <= L && r >= R) {
    return tree[rt];
  }
  int mid = (L + R) >> 1;
  push_up(tree[2 * rt], tree[2 * rt + 1]);
  node X, Y;
  if (mid < l) return query(2 * rt + 1, mid + 1, R, l, r);
  if (mid >= r) return query(2 * rt, L, mid, l, r);
  return query(2 * rt, L, mid, l, r) + query(2 * rt + 1, mid + 1, R, l, r);
}
int main() {
  cin >> n >> q;
  scanf("%s", num + 1);
  build_tree(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.val[0][4] == INF) {
      ans.val[0][4] = -1;
    }
    printf("%d\n", ans.val[0][4]);
  }
  return 0;
}

### Prompt

Please provide a cpp coded solution to the problem described below:

A string t is called nice if a string "2017" occurs in t as a subsequence but a string "2016" doesn't occur in t as a subsequence. For example, strings "203434107" and "9220617" are nice, while strings "20016", "1234" and "20167" aren't nice.

The ugliness of a string is the minimum possible number of characters to remove, in order to obtain a nice string. If it's impossible to make a string nice by removing characters, its ugliness is  - 1.

Limak has a string s of length n, with characters indexed 1 through n. He asks you q queries. In the i-th query you should compute and print the ugliness of a substring (continuous subsequence) of s starting at the index ai and ending at the index bi (inclusive).

Input

The first line of the input contains two integers n and q (4 ≤ n ≤ 200 000, 1 ≤ q ≤ 200 000) — the length of the string s and the number of queries respectively.

The second line contains a string s of length n. Every character is one of digits '0'–'9'.

The i-th of next q lines contains two integers ai and bi (1 ≤ ai ≤ bi ≤ n), describing a substring in the i-th query.

Output

For each query print the ugliness of the given substring.

Examples

Input

8 3
20166766
1 8
1 7
2 8


Output

4
3
-1


Input

15 5
012016662091670
3 4
1 14
4 15
1 13
10 15


Output

-1
2
1
-1
-1


Input

4 2
1234
2 4
1 2


Output

-1
-1

Note

In the first sample:

  * In the first query, ugliness("20166766") = 4 because all four sixes must be removed. 
  * In the second query, ugliness("2016676") = 3 because all three sixes must be removed. 
  * In the third query, ugliness("0166766") = - 1 because it's impossible to remove some digits to get a nice string. 



In the second sample:

  * In the second query, ugliness("01201666209167") = 2. It's optimal to remove the first digit '2' and the last digit '6', what gives a string "010166620917", which is nice. 
  * In the third query, ugliness("016662091670") = 1. It's optimal to remove the last digit '6', what gives a nice string "01666209170". 

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int maxn = 200010;
struct node {
  int val[5][5];
  node() {
    for (int i = 0; i < 5; i++) {
      for (int j = 0; j < 5; j++) {
        val[i][j] = INF;
      }
    }
  }
  node operator+(node x) {
    node ans;
    for (int i = 0; i < 5; ++i) {
      for (int j = 0; j < 5; ++j) {
        for (int k = 0; k < 5; ++k) {
          ans.val[i][j] = min(ans.val[i][j], val[i][k] + x.val[k][j]);
        }
      }
    }
    return ans;
  }
} tree[maxn << 2];
int n, q;
char num[maxn];
char ch[] = {'2', '0', '1', '7', '6'};
int Find(char c) {
  for (int i = 0; i < 5; i++) {
    if (ch[i] == c) return i;
  }
  return -1;
}
node push_up(node &a, node &b) {
  node res;
  for (int i = 0; i < 5; i++) {
    for (int j = i; j < 5; j++) {
      for (int k = i; k <= j; k++) {
        res.val[i][j] = min(res.val[i][j], a.val[i][k] + b.val[k][j]);
      }
    }
  }
  return res;
}
void build_tree(int rt, int l, int r) {
  if (l == r) {
    int f = Find(num[l]);
    for (int i = 0; i < 5; i++) {
      tree[rt].val[i][i] = 0;
    }
    if (f != -1 && f < 4) {
      tree[rt].val[f][f + 1] = 0;
      tree[rt].val[f][f] = 1;
    } else if (f == 4) {
      tree[rt].val[3][3] = tree[rt].val[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) >> 1;
  build_tree(rt << 1, l, mid);
  build_tree(rt << 1 | 1, mid + 1, r);
  tree[rt] = push_up(tree[rt << 1], tree[rt << 1 | 1]);
}
node query(int rt, int L, int R, int l, int r) {
  if (l <= L && r >= R) {
    return tree[rt];
  }
  int mid = (L + R) >> 1;
  push_up(tree[2 * rt], tree[2 * rt + 1]);
  node X, Y;
  if (mid < l) return query(2 * rt + 1, mid + 1, R, l, r);
  if (mid >= r) return query(2 * rt, L, mid, l, r);
  return query(2 * rt, L, mid, l, r) + query(2 * rt + 1, mid + 1, R, l, r);
}
int main() {
  cin >> n >> q;
  scanf("%s", num + 1);
  build_tree(1, 1, n);
  while (q--) {
    int l, r;
    scanf("%d %d", &l, &r);
    node ans = query(1, 1, n, l, r);
    if (ans.val[0][4] == INF) {
      ans.val[0][4] = -1;
    }
    printf("%d\n", ans.val[0][4]);
  }
  return 0;
}

```