#include <bits/stdc++.h>
using namespace std;
const int Max_N = 2e5;
int n, q;
char a[Max_N + 10] = {};
struct kk {
  int l, r, m[7][6];
} A[Max_N * 4 + 10] = {};
int m1[7][7] = {}, m2[7][7] = {}, m3[7][7] = {};
inline void Mull() {
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 7; j++) m3[i][j] = 1 << 20;
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 7; j++)
      for (int k = 0; k < 7; k++) m3[i][k] = min(m3[i][k], m1[i][j] + m2[j][k]);
}
inline void Build(int x, int l, int r) {
  A[x].l = l;
  A[x].r = r;
  if (l == r) {
    for (int i = 0; i < 7; i++)
      for (int j = 0; j < 6; j++) A[x].m[i][j] = 1 << 20;
    for (int i = 0; i < 6; i++) A[x].m[i][i] = 0;
    if (a[l] == '2') A[x].m[0][1] = 0, A[x].m[0][0] = 1;
    if (a[l] == '0') A[x].m[1][2] = 0, A[x].m[1][1] = 1;
    if (a[l] == '1') A[x].m[2][3] = 0, A[x].m[2][2] = 1;
    if (a[l] == '7') A[x].m[3][4] = 0, A[x].m[3][3] = 1;
    if (a[l] == '6')
      A[x].m[3][5] = A[x].m[4][5] = 0, A[x].m[3][3] = A[x].m[4][4] = 1;
    return;
  }
  Build(x << 1, l, (l + r) >> 1);
  Build((x << 1) + 1, ((l + r) >> 1) + 1, r);
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 6; j++)
      m1[i][j] = A[x << 1].m[i][j], m2[i][j] = A[(x << 1) + 1].m[i][j];
  Mull();
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 6; j++) A[x].m[i][j] = m3[i][j];
}
int f[6] = {}, res[6] = {};
int vis = 0, M[7][7] = {};
inline void Mul() {
  for (int i = 0; i < 5; i++) res[i] = f[i], f[i] = 1 << 30;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) f[j] = min(f[j], res[i] + M[i][j]);
  }
}
inline void Ask(int x, int l, int r) {
  if (l <= A[x].l && A[x].r <= r) {
    if (!vis) {
      vis = 1;
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) M[i][j] = A[x].m[i][j];
    } else {
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) m1[i][j] = M[i][j], m2[i][j] = A[x].m[i][j];
      Mull();
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) M[i][j] = m3[i][j];
    }
    return;
  }
  if (l <= A[x << 1].r) Ask(x << 1, l, r);
  if (r > A[x << 1].r) Ask((x << 1) + 1, l, r);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", a + 1);
  Build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    f[0] = 0;
    for (int j = 1; j < 6; j++) f[j] = 1 << 20;
    for (int j = 0; j < 7; j++)
      for (int k = 0; k < 7; k++) M[j][k] = 1 << 20;
    vis = 0;
    Ask(1, l, r);
    Mul();
    if (f[4] >= n)
      puts("-1");
    else
      printf("%d\n", f[4]);
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 Max_N = 2e5;
int n, q;
char a[Max_N + 10] = {};
struct kk {
  int l, r, m[7][6];
} A[Max_N * 4 + 10] = {};
int m1[7][7] = {}, m2[7][7] = {}, m3[7][7] = {};
inline void Mull() {
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 7; j++) m3[i][j] = 1 << 20;
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 7; j++)
      for (int k = 0; k < 7; k++) m3[i][k] = min(m3[i][k], m1[i][j] + m2[j][k]);
}
inline void Build(int x, int l, int r) {
  A[x].l = l;
  A[x].r = r;
  if (l == r) {
    for (int i = 0; i < 7; i++)
      for (int j = 0; j < 6; j++) A[x].m[i][j] = 1 << 20;
    for (int i = 0; i < 6; i++) A[x].m[i][i] = 0;
    if (a[l] == '2') A[x].m[0][1] = 0, A[x].m[0][0] = 1;
    if (a[l] == '0') A[x].m[1][2] = 0, A[x].m[1][1] = 1;
    if (a[l] == '1') A[x].m[2][3] = 0, A[x].m[2][2] = 1;
    if (a[l] == '7') A[x].m[3][4] = 0, A[x].m[3][3] = 1;
    if (a[l] == '6')
      A[x].m[3][5] = A[x].m[4][5] = 0, A[x].m[3][3] = A[x].m[4][4] = 1;
    return;
  }
  Build(x << 1, l, (l + r) >> 1);
  Build((x << 1) + 1, ((l + r) >> 1) + 1, r);
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 6; j++)
      m1[i][j] = A[x << 1].m[i][j], m2[i][j] = A[(x << 1) + 1].m[i][j];
  Mull();
  for (int i = 0; i < 7; i++)
    for (int j = 0; j < 6; j++) A[x].m[i][j] = m3[i][j];
}
int f[6] = {}, res[6] = {};
int vis = 0, M[7][7] = {};
inline void Mul() {
  for (int i = 0; i < 5; i++) res[i] = f[i], f[i] = 1 << 30;
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) f[j] = min(f[j], res[i] + M[i][j]);
  }
}
inline void Ask(int x, int l, int r) {
  if (l <= A[x].l && A[x].r <= r) {
    if (!vis) {
      vis = 1;
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) M[i][j] = A[x].m[i][j];
    } else {
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) m1[i][j] = M[i][j], m2[i][j] = A[x].m[i][j];
      Mull();
      for (int i = 0; i < 7; i++)
        for (int j = 0; j < 7; j++) M[i][j] = m3[i][j];
    }
    return;
  }
  if (l <= A[x << 1].r) Ask(x << 1, l, r);
  if (r > A[x << 1].r) Ask((x << 1) + 1, l, r);
}
int main() {
  scanf("%d%d", &n, &q);
  scanf("%s", a + 1);
  Build(1, 1, n);
  for (int i = 1; i <= q; i++) {
    int l, r;
    scanf("%d%d", &l, &r);
    f[0] = 0;
    for (int j = 1; j < 6; j++) f[j] = 1 << 20;
    for (int j = 0; j < 7; j++)
      for (int k = 0; k < 7; k++) M[j][k] = 1 << 20;
    vis = 0;
    Ask(1, l, r);
    Mul();
    if (f[4] >= n)
      puts("-1");
    else
      printf("%d\n", f[4]);
  }
  return 0;
}

```