#include <bits/stdc++.h>
using namespace std;
#pragma comment(linker, "/STACK:102400000,102400000")
const int inf = 0x3f3f3f3f;
const int maxn = 200000 + 5;
struct Node {
  int dp[5][5];
} tree[maxn << 2];
int n, q;
char ch[maxn];
map<char, int> id;
inline void init(Node &t) {
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.dp[i][j] = inf;
}
Node Merge(Node a, Node b) {
  Node tmp;
  init(tmp);
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++)
      for (int k = i; k <= j; k++)
        tmp.dp[i][j] = min(tmp.dp[i][j], a.dp[i][k] + b.dp[k][j]);
  return tmp;
}
void display(Node t) {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) printf("%d ", t.dp[i][j]);
    cout << endl;
  }
}
void buildtree(int l, int r, int o) {
  if (l == r) {
    init(tree[o]);
    for (int i = 0; i < 5; i++) tree[o].dp[i][i] = 0;
    int pos = id[ch[l]];
    if (pos <= 3) tree[o].dp[pos][pos] = 1, tree[o].dp[pos][pos + 1] = 0;
    if (pos == 4) tree[o].dp[3][3] = tree[o].dp[4][4] = 1;
    return;
  }
  int mid = (l + r) / 2;
  if (l <= mid) buildtree(l, mid, o << 1);
  if (r > mid) buildtree(mid + 1, r, o << 1 | 1);
  tree[o] = Merge(tree[o << 1], tree[o << 1 | 1]);
}
Node ans;
void query(int ql, int qr, int l, int r, int o) {
  if (ql <= l && qr >= r) {
    if (ql == l)
      ans = tree[o];
    else
      ans = Merge(ans, tree[o]);
    return;
  }
  int mid = (l + r) / 2;
  if (ql <= mid) query(ql, qr, l, mid, o << 1);
  if (qr > mid) query(ql, qr, mid + 1, r, o << 1 | 1);
}
int main() {
  int cnt = 4;
  id['2'] = 0, id['0'] = 1, id['1'] = 2, id['7'] = 3, id['6'] = 4;
  for (char i = '3'; i <= '9'; i++)
    if (id.count(i) == 0) id[i] = ++cnt;
  cin >> n >> q;
  scanf("%s", ch + 1);
  buildtree(1, n, 1);
  for (int i = 1; i <= q; i++) {
    int ql, qr;
    scanf("%d%d", &ql, &qr);
    init(ans);
    query(ql, qr, 1, n, 1);
    if (ans.dp[0][4] >= inf)
      puts("-1");
    else
      printf("%d\n", ans.dp[0][4]);
  }
  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;
#pragma comment(linker, "/STACK:102400000,102400000")
const int inf = 0x3f3f3f3f;
const int maxn = 200000 + 5;
struct Node {
  int dp[5][5];
} tree[maxn << 2];
int n, q;
char ch[maxn];
map<char, int> id;
inline void init(Node &t) {
  for (int i = 0; i < 5; i++)
    for (int j = 0; j < 5; j++) t.dp[i][j] = inf;
}
Node Merge(Node a, Node b) {
  Node tmp;
  init(tmp);
  for (int i = 0; i <= 4; i++)
    for (int j = i; j <= 4; j++)
      for (int k = i; k <= j; k++)
        tmp.dp[i][j] = min(tmp.dp[i][j], a.dp[i][k] + b.dp[k][j]);
  return tmp;
}
void display(Node t) {
  for (int i = 0; i < 5; i++) {
    for (int j = 0; j < 5; j++) printf("%d ", t.dp[i][j]);
    cout << endl;
  }
}
void buildtree(int l, int r, int o) {
  if (l == r) {
    init(tree[o]);
    for (int i = 0; i < 5; i++) tree[o].dp[i][i] = 0;
    int pos = id[ch[l]];
    if (pos <= 3) tree[o].dp[pos][pos] = 1, tree[o].dp[pos][pos + 1] = 0;
    if (pos == 4) tree[o].dp[3][3] = tree[o].dp[4][4] = 1;
    return;
  }
  int mid = (l + r) / 2;
  if (l <= mid) buildtree(l, mid, o << 1);
  if (r > mid) buildtree(mid + 1, r, o << 1 | 1);
  tree[o] = Merge(tree[o << 1], tree[o << 1 | 1]);
}
Node ans;
void query(int ql, int qr, int l, int r, int o) {
  if (ql <= l && qr >= r) {
    if (ql == l)
      ans = tree[o];
    else
      ans = Merge(ans, tree[o]);
    return;
  }
  int mid = (l + r) / 2;
  if (ql <= mid) query(ql, qr, l, mid, o << 1);
  if (qr > mid) query(ql, qr, mid + 1, r, o << 1 | 1);
}
int main() {
  int cnt = 4;
  id['2'] = 0, id['0'] = 1, id['1'] = 2, id['7'] = 3, id['6'] = 4;
  for (char i = '3'; i <= '9'; i++)
    if (id.count(i) == 0) id[i] = ++cnt;
  cin >> n >> q;
  scanf("%s", ch + 1);
  buildtree(1, n, 1);
  for (int i = 1; i <= q; i++) {
    int ql, qr;
    scanf("%d%d", &ql, &qr);
    init(ans);
    query(ql, qr, 1, n, 1);
    if (ans.dp[0][4] >= inf)
      puts("-1");
    else
      printf("%d\n", ans.dp[0][4]);
  }
  return 0;
}

```