#include <bits/stdc++.h>
using namespace std;
const int INF = 1e9;
const int maxn = 2e5 + 10;
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 * 4];
char ch[10] = {'2', '0', '1', '7', '6'};
char s[maxn];
node pushup(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;
}
int Find(char c) {
  for (int i = 0; i < 5; i++) {
    if (ch[i] == c) return i;
  }
  return -1;
}
void build(int l, int r, int rt) {
  if (l == r) {
    int t = Find(s[l]);
    for (int i = 0; i < 5; i++) {
      tree[rt].val[i][i] = 0;
    }
    if (t != -1 && t < 4) {
      tree[rt].val[t][t + 1] = 0;
      tree[rt].val[t][t] = 1;
    } else if (t == 4) {
      tree[rt].val[3][3] = tree[rt].val[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * rt);
  build(mid + 1, r, 2 * rt + 1);
  tree[rt] = pushup(tree[2 * rt], tree[2 * rt + 1]);
}
node query(int l, int r, int rt, int L, int R) {
  if (L <= l && R >= r) {
    return tree[rt];
  }
  int mid = (l + r) / 2;
  if (mid < L) return query(mid + 1, r, 2 * rt + 1, L, R);
  if (mid >= R) return query(l, mid, 2 * rt, L, R);
  return query(l, mid, 2 * rt, L, R) + query(mid + 1, r, 2 * rt + 1, L, R);
}
int main() {
  int n, m, l, r;
  scanf("%d%d%s", &n, &m, s + 1);
  build(1, n, 1);
  while (m--) {
    node ans;
    scanf("%d%d", &l, &r);
    ans = query(1, n, 1, l, r);
    if (ans.val[0][4] == INF)
      printf("-1\n");
    else
      printf("%d\n", ans.val[0][4]);
  }
  return 0;
}

### Prompt

Generate a CPP solution to the following problem:

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 = 2e5 + 10;
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 * 4];
char ch[10] = {'2', '0', '1', '7', '6'};
char s[maxn];
node pushup(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;
}
int Find(char c) {
  for (int i = 0; i < 5; i++) {
    if (ch[i] == c) return i;
  }
  return -1;
}
void build(int l, int r, int rt) {
  if (l == r) {
    int t = Find(s[l]);
    for (int i = 0; i < 5; i++) {
      tree[rt].val[i][i] = 0;
    }
    if (t != -1 && t < 4) {
      tree[rt].val[t][t + 1] = 0;
      tree[rt].val[t][t] = 1;
    } else if (t == 4) {
      tree[rt].val[3][3] = tree[rt].val[4][4] = 1;
    }
    return;
  }
  int mid = (l + r) / 2;
  build(l, mid, 2 * rt);
  build(mid + 1, r, 2 * rt + 1);
  tree[rt] = pushup(tree[2 * rt], tree[2 * rt + 1]);
}
node query(int l, int r, int rt, int L, int R) {
  if (L <= l && R >= r) {
    return tree[rt];
  }
  int mid = (l + r) / 2;
  if (mid < L) return query(mid + 1, r, 2 * rt + 1, L, R);
  if (mid >= R) return query(l, mid, 2 * rt, L, R);
  return query(l, mid, 2 * rt, L, R) + query(mid + 1, r, 2 * rt + 1, L, R);
}
int main() {
  int n, m, l, r;
  scanf("%d%d%s", &n, &m, s + 1);
  build(1, n, 1);
  while (m--) {
    node ans;
    scanf("%d%d", &l, &r);
    ans = query(1, n, 1, l, r);
    if (ans.val[0][4] == INF)
      printf("-1\n");
    else
      printf("%d\n", ans.val[0][4]);
  }
  return 0;
}

```