#include <bits/stdc++.h>
using namespace std;
struct node {
  int a[6][6];
};
char buf[200010];
node tree[2 * (1 << 18)];
int pre[] = {2, 0, 1, 7, 6};
void init_node(int id, int x) {
  int i;
  for (int i = 0; i <= 4; i++) tree[id].a[i][i] = 1;
  for ((i) = 0; (i) < (int)(5); (i)++)
    if (i != 3) {
      if (x == pre[i]) {
        tree[id].a[i][i + 1] = 0;
      } else {
        tree[id].a[i][i] = 0;
      }
    }
  if (x == 6) {
    tree[id].a[3][5] = (1 << 29);
  } else if (x == 7) {
    tree[id].a[3][4] = 0;
  } else {
    tree[id].a[3][3] = 0;
  }
}
node merge(node &x, node &y) {
  node z;
  int i, j, k;
  for ((i) = 0; (i) < (int)(5); (i)++)
    for ((j) = 0; (j) < (int)(5); (j)++) z.a[i][j] = (1 << 29);
  for ((i) = 0; (i) < (int)(5); (i)++)
    for (j = i; j < 5; j++)
      for (k = j; k < 5; k++) z.a[i][k] = min(z.a[i][k], x.a[i][j] + y.a[j][k]);
  return z;
}
node query(int L, int R, int id = 1, int low = 0, int high = (1 << 18)) {
  if (L <= low && R >= high) return tree[id];
  int mid = (low + high) / 2;
  if (R <= mid) {
    return query(L, R, 2 * id, low, mid);
  } else if (L >= mid) {
    return query(L, R, 2 * id + 1, mid, high);
  } else {
    node x = query(L, R, 2 * id, low, mid);
    node y = query(L, R, 2 * id + 1, mid, high);
    return merge(x, y);
  }
}
int main(void) {
  int N, Q, i, j, k;
  cin >> N >> Q;
  scanf("%s", buf);
  for ((i) = 0; (i) < (int)(2 * (1 << 18)); (i)++)
    for ((j) = 0; (j) < (int)(5); (j)++)
      for ((k) = 0; (k) < (int)(5); (k)++) tree[i].a[j][k] = (1 << 29);
  for ((i) = 0; (i) < (int)(N); (i)++) init_node((1 << 18) + i, buf[i] - '0');
  for (i = (1 << 18) - 1; i >= 1; i--) {
    if (2 * i + 1 <= (1 << 18) + N - 1) i++, i--;
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    int L, R;
    scanf("%d%d", &L, &R);
    L--;
    node ans = query(L, R);
    int tmp = ans.a[0][4];
    if (tmp == (1 << 29)) tmp = -1;
    printf("%d\n", tmp);
  }
  return 0;
}

### Prompt

Your task is to create 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;
struct node {
  int a[6][6];
};
char buf[200010];
node tree[2 * (1 << 18)];
int pre[] = {2, 0, 1, 7, 6};
void init_node(int id, int x) {
  int i;
  for (int i = 0; i <= 4; i++) tree[id].a[i][i] = 1;
  for ((i) = 0; (i) < (int)(5); (i)++)
    if (i != 3) {
      if (x == pre[i]) {
        tree[id].a[i][i + 1] = 0;
      } else {
        tree[id].a[i][i] = 0;
      }
    }
  if (x == 6) {
    tree[id].a[3][5] = (1 << 29);
  } else if (x == 7) {
    tree[id].a[3][4] = 0;
  } else {
    tree[id].a[3][3] = 0;
  }
}
node merge(node &x, node &y) {
  node z;
  int i, j, k;
  for ((i) = 0; (i) < (int)(5); (i)++)
    for ((j) = 0; (j) < (int)(5); (j)++) z.a[i][j] = (1 << 29);
  for ((i) = 0; (i) < (int)(5); (i)++)
    for (j = i; j < 5; j++)
      for (k = j; k < 5; k++) z.a[i][k] = min(z.a[i][k], x.a[i][j] + y.a[j][k]);
  return z;
}
node query(int L, int R, int id = 1, int low = 0, int high = (1 << 18)) {
  if (L <= low && R >= high) return tree[id];
  int mid = (low + high) / 2;
  if (R <= mid) {
    return query(L, R, 2 * id, low, mid);
  } else if (L >= mid) {
    return query(L, R, 2 * id + 1, mid, high);
  } else {
    node x = query(L, R, 2 * id, low, mid);
    node y = query(L, R, 2 * id + 1, mid, high);
    return merge(x, y);
  }
}
int main(void) {
  int N, Q, i, j, k;
  cin >> N >> Q;
  scanf("%s", buf);
  for ((i) = 0; (i) < (int)(2 * (1 << 18)); (i)++)
    for ((j) = 0; (j) < (int)(5); (j)++)
      for ((k) = 0; (k) < (int)(5); (k)++) tree[i].a[j][k] = (1 << 29);
  for ((i) = 0; (i) < (int)(N); (i)++) init_node((1 << 18) + i, buf[i] - '0');
  for (i = (1 << 18) - 1; i >= 1; i--) {
    if (2 * i + 1 <= (1 << 18) + N - 1) i++, i--;
    tree[i] = merge(tree[2 * i], tree[2 * i + 1]);
  }
  for ((i) = 0; (i) < (int)(Q); (i)++) {
    int L, R;
    scanf("%d%d", &L, &R);
    L--;
    node ans = query(L, R);
    int tmp = ans.a[0][4];
    if (tmp == (1 << 29)) tmp = -1;
    printf("%d\n", tmp);
  }
  return 0;
}

```