#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int r = power(a, b / 2) % 1000000007;
  if (b % 2 == 0)
    return (r * r) % 1000000007;
  else
    return (((r * r) % 1000000007) * a) % 1000000007;
}
int prime[1000009];
int sumprime[1000009];
void sieve(int n) {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= n; j = j + i) {
        prime[j]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!prime[i]) {
      sumprime[i] = sumprime[i - 1] + 1;
    } else {
      sumprime[i] = sumprime[i - 1];
    }
  }
}
bool comp(const pair<long long int, long long int> &a,
          const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
const int N = 5 * 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[n], b[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
  }
  map<long long int, long long int> mp;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 10000; j++) {
      if (j - b[i] >= 0 && (j - b[i]) % a[i] == 0) {
        if (s[i] == '1') {
          s[i] = '0';
        } else {
          s[i] = '1';
        }
      }
      if (s[i] == '1') {
        mp[j]++;
      }
    }
  }
  long long int ma = 0;
  for (auto it : mp) {
    ma = max(ma, it.second);
  }
  cout << ma << "\n";
  return 0;
}

### Prompt

Generate a cpp solution to the following problem:

It is a holiday season, and Koala is decorating his house with cool lights! He owns n lights, all of which flash periodically.

After taking a quick glance at them, Koala realizes that each of his lights can be described with two parameters a_i and b_i. Light with parameters a_i and b_i will toggle (on to off, or off to on) every a_i seconds starting from the b_i-th second. In other words, it will toggle at the moments b_i, b_i + a_i, b_i + 2 ⋅ a_i and so on.

You know for each light whether it's initially on or off and its corresponding parameters a_i and b_i. Koala is wondering what is the maximum number of lights that will ever be on at the same time. So you need to find that out.

<image> Here is a graphic for the first example.

Input

The first line contains a single integer n (1 ≤ n ≤ 100), the number of lights.

The next line contains a string s of n characters. The i-th character is "1", if the i-th lamp is initially on. Otherwise, i-th character is "0".

The i-th of the following n lines contains two integers a_i and b_i (1 ≤ a_i, b_i ≤ 5) — the parameters of the i-th light.

Output

Print a single integer — the maximum number of lights that will ever be on at the same time.

Examples

Input


3
101
3 3
3 2
3 1


Output


2


Input


4
1111
3 4
5 2
3 1
3 2


Output


4


Input


6
011100
5 3
5 5
2 4
3 5
4 2
1 5


Output


6

Note

For first example, the lamps' states are shown in the picture above. The largest number of simultaneously on lamps is 2 (e.g. at the moment 2).

In the second example, all lights are initially on. So the answer is 4.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long int power(long long int a, long long int b) {
  if (b == 0) return 1;
  if (b == 1) return a;
  long long int r = power(a, b / 2) % 1000000007;
  if (b % 2 == 0)
    return (r * r) % 1000000007;
  else
    return (((r * r) % 1000000007) * a) % 1000000007;
}
int prime[1000009];
int sumprime[1000009];
void sieve(int n) {
  prime[0] = prime[1] = 1;
  for (int i = 2; i <= sqrt(n); i++) {
    if (!prime[i]) {
      for (int j = 2 * i; j <= n; j = j + i) {
        prime[j]++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (!prime[i]) {
      sumprime[i] = sumprime[i - 1] + 1;
    } else {
      sumprime[i] = sumprime[i - 1];
    }
  }
}
bool comp(const pair<long long int, long long int> &a,
          const pair<long long int, long long int> &b) {
  return (a.second < b.second);
}
const int N = 5 * 1e5 + 9;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  string s;
  cin >> s;
  int a[n], b[n];
  for (int i = 0; i < n; i++) {
    cin >> a[i] >> b[i];
  }
  map<long long int, long long int> mp;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 10000; j++) {
      if (j - b[i] >= 0 && (j - b[i]) % a[i] == 0) {
        if (s[i] == '1') {
          s[i] = '0';
        } else {
          s[i] = '1';
        }
      }
      if (s[i] == '1') {
        mp[j]++;
      }
    }
  }
  long long int ma = 0;
  for (auto it : mp) {
    ma = max(ma, it.second);
  }
  cout << ma << "\n";
  return 0;
}

```