#include <bits/stdc++.h>
using namespace std;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n, m;
char s[200002];
struct mat {
  int a[6][6];
  mat() { memset(a, 0, sizeof(a)); }
  void init() {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        a[i][j] = i == j ? 0 : 0x3f3f3f3f;
  }
  mat operator+(const mat &b) const {
    mat res;
    memset(res.a, 0x3f3f3f3f, sizeof(res.a));
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        for (int k = (0), k_end_ = (4); k <= k_end_; k++)
          chkmin(res.a[i][j], a[i][k] + b.a[k][j]);
    return res;
  }
  void Pr() {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        printf("%d%c", a[i][j], j < 4 ? ' ' : '\n');
  }
};
struct Tree {
  struct node {
    int L, R;
    mat sum;
  } tree[200002 << 2];
  void Up(node &A, node &L, node &R) { A.sum = L.sum + R.sum; }
  void build(int L, int R, int p) {
    tree[p].L = L, tree[p].R = R;
    if (L == R) {
      tree[p].sum.init();
      if (s[L] == '2') tree[p].sum.a[0][0] = 1, tree[p].sum.a[0][1] = 0;
      if (s[L] == '0') tree[p].sum.a[1][1] = 1, tree[p].sum.a[1][2] = 0;
      if (s[L] == '1') tree[p].sum.a[2][2] = 1, tree[p].sum.a[2][3] = 0;
      if (s[L] == '7') tree[p].sum.a[3][3] = 1, tree[p].sum.a[3][4] = 0;
      if (s[L] == '6') tree[p].sum.a[3][3] = 1, tree[p].sum.a[4][4] = 1;
      return;
    }
    int mid = (L + R) >> 1;
    build(L, mid, p << 1), build(mid + 1, R, p << 1 | 1);
    Up(tree[p], tree[p << 1], tree[p << 1 | 1]);
  }
  mat query(int L, int R, int p) {
    if (tree[p].L == L && tree[p].R == R) return tree[p].sum;
    int mid = (tree[p].L + tree[p].R) >> 1;
    if (R <= mid)
      return query(L, R, p << 1);
    else if (L > mid)
      return query(L, R, p << 1 | 1);
    else
      return query(L, mid, p << 1) + query(mid + 1, R, p << 1 | 1);
  }
} T;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  T.build(1, n, 1);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    mat ans = T.query(l, r, 1);
    printf("%d\n", ans.a[0][4] == 0x3f3f3f3f ? -1 : ans.a[0][4]);
  }
  return 0;
}

### Prompt

Please formulate 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;
template <class T>
inline bool chkmin(T &x, T y) {
  return y < x ? x = y, 1 : 0;
}
template <class T>
inline bool chkmax(T &x, T y) {
  return x < y ? x = y, 1 : 0;
}
inline long long Max(long long x, long long y) { return x > y ? x : y; }
inline long long Min(long long x, long long y) { return x < y ? x : y; }
int n, m;
char s[200002];
struct mat {
  int a[6][6];
  mat() { memset(a, 0, sizeof(a)); }
  void init() {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        a[i][j] = i == j ? 0 : 0x3f3f3f3f;
  }
  mat operator+(const mat &b) const {
    mat res;
    memset(res.a, 0x3f3f3f3f, sizeof(res.a));
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        for (int k = (0), k_end_ = (4); k <= k_end_; k++)
          chkmin(res.a[i][j], a[i][k] + b.a[k][j]);
    return res;
  }
  void Pr() {
    for (int i = (0), i_end_ = (4); i <= i_end_; i++)
      for (int j = (0), j_end_ = (4); j <= j_end_; j++)
        printf("%d%c", a[i][j], j < 4 ? ' ' : '\n');
  }
};
struct Tree {
  struct node {
    int L, R;
    mat sum;
  } tree[200002 << 2];
  void Up(node &A, node &L, node &R) { A.sum = L.sum + R.sum; }
  void build(int L, int R, int p) {
    tree[p].L = L, tree[p].R = R;
    if (L == R) {
      tree[p].sum.init();
      if (s[L] == '2') tree[p].sum.a[0][0] = 1, tree[p].sum.a[0][1] = 0;
      if (s[L] == '0') tree[p].sum.a[1][1] = 1, tree[p].sum.a[1][2] = 0;
      if (s[L] == '1') tree[p].sum.a[2][2] = 1, tree[p].sum.a[2][3] = 0;
      if (s[L] == '7') tree[p].sum.a[3][3] = 1, tree[p].sum.a[3][4] = 0;
      if (s[L] == '6') tree[p].sum.a[3][3] = 1, tree[p].sum.a[4][4] = 1;
      return;
    }
    int mid = (L + R) >> 1;
    build(L, mid, p << 1), build(mid + 1, R, p << 1 | 1);
    Up(tree[p], tree[p << 1], tree[p << 1 | 1]);
  }
  mat query(int L, int R, int p) {
    if (tree[p].L == L && tree[p].R == R) return tree[p].sum;
    int mid = (tree[p].L + tree[p].R) >> 1;
    if (R <= mid)
      return query(L, R, p << 1);
    else if (L > mid)
      return query(L, R, p << 1 | 1);
    else
      return query(L, mid, p << 1) + query(mid + 1, R, p << 1 | 1);
  }
} T;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  T.build(1, n, 1);
  while (m--) {
    int l, r;
    scanf("%d%d", &l, &r);
    mat ans = T.query(l, r, 1);
    printf("%d\n", ans.a[0][4] == 0x3f3f3f3f ? -1 : ans.a[0][4]);
  }
  return 0;
}

```