#include <bits/stdc++.h>
using namespace std;
char s[200005];
int n, q;
struct node {
  int v[5][5];
} st[4 * 200005];
node getv(char c) {
  node x;
  int i, j, k;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) x.v[i][j] = 1e9;
    x.v[i][i] = 0;
  }
  if (c == '2')
    x.v[0][0] = 1, x.v[0][1] = 0;
  else if (c == '0')
    x.v[1][1] = 1, x.v[1][2] = 0;
  else if (c == '1')
    x.v[2][2] = 1, x.v[2][3] = 0;
  else if (c == '6')
    x.v[3][3] = 1, x.v[4][4] = 1;
  else if (c == '7')
    x.v[3][3] = 1, x.v[3][4] = 0;
  return x;
}
node merge(node l, node r) {
  node x;
  int i, j, k;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++) x.v[i][j] = 1e9;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      for (k = 0; k < 5; k++) x.v[i][j] = min(x.v[i][j], l.v[i][k] + r.v[k][j]);
  return x;
}
void build(int cur, int L, int R) {
  if (L == R) {
    st[cur] = getv(s[L]);
    return;
  }
  build((cur << 1), L, ((L + R) / 2));
  build(((cur << 1) | 1), ((L + R) / 2) + 1, R);
  st[cur] = merge(st[(cur << 1)], st[((cur << 1) | 1)]);
}
node query(int cur, int L, int R, int i, int j) {
  if (i > R || j < L) return getv('9');
  if (i <= L && R <= j) return st[cur];
  return merge(query((cur << 1), L, ((L + R) / 2), i, j),
               query(((cur << 1) | 1), ((L + R) / 2) + 1, R, i, j));
}
int main() {
  int i, j, k, l, r;
  scanf(" %d %d", &n, &q);
  scanf(" %s", s);
  build(1, 0, n - 1);
  while (q--) {
    scanf(" %d %d", &l, &r);
    l--;
    r--;
    int res = query(1, 0, n - 1, l, r).v[0][4];
    if (res >= 1e9)
      printf("-1\n");
    else
      printf("%d\n", res);
  }
  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;
char s[200005];
int n, q;
struct node {
  int v[5][5];
} st[4 * 200005];
node getv(char c) {
  node x;
  int i, j, k;
  for (i = 0; i < 5; i++) {
    for (j = 0; j < 5; j++) x.v[i][j] = 1e9;
    x.v[i][i] = 0;
  }
  if (c == '2')
    x.v[0][0] = 1, x.v[0][1] = 0;
  else if (c == '0')
    x.v[1][1] = 1, x.v[1][2] = 0;
  else if (c == '1')
    x.v[2][2] = 1, x.v[2][3] = 0;
  else if (c == '6')
    x.v[3][3] = 1, x.v[4][4] = 1;
  else if (c == '7')
    x.v[3][3] = 1, x.v[3][4] = 0;
  return x;
}
node merge(node l, node r) {
  node x;
  int i, j, k;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++) x.v[i][j] = 1e9;
  for (i = 0; i < 5; i++)
    for (j = 0; j < 5; j++)
      for (k = 0; k < 5; k++) x.v[i][j] = min(x.v[i][j], l.v[i][k] + r.v[k][j]);
  return x;
}
void build(int cur, int L, int R) {
  if (L == R) {
    st[cur] = getv(s[L]);
    return;
  }
  build((cur << 1), L, ((L + R) / 2));
  build(((cur << 1) | 1), ((L + R) / 2) + 1, R);
  st[cur] = merge(st[(cur << 1)], st[((cur << 1) | 1)]);
}
node query(int cur, int L, int R, int i, int j) {
  if (i > R || j < L) return getv('9');
  if (i <= L && R <= j) return st[cur];
  return merge(query((cur << 1), L, ((L + R) / 2), i, j),
               query(((cur << 1) | 1), ((L + R) / 2) + 1, R, i, j));
}
int main() {
  int i, j, k, l, r;
  scanf(" %d %d", &n, &q);
  scanf(" %s", s);
  build(1, 0, n - 1);
  while (q--) {
    scanf(" %d %d", &l, &r);
    l--;
    r--;
    int res = query(1, 0, n - 1, l, r).v[0][4];
    if (res >= 1e9)
      printf("-1\n");
    else
      printf("%d\n", res);
  }
  return 0;
}

```