#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
long long add(long long x, long long y, long long CMOD = MOD) {
  return (x + y + CMOD + CMOD) % CMOD;
}
long long mult(long long x, long long y, long long CMOD = MOD) {
  return add(x, 0) * add(y, 0) % CMOD;
}
long long fast_expo(long long x, long long y, long long CMOD = MOD) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long ans = fast_expo(x, y / 2, CMOD);
  ans = mult(ans, ans, CMOD);
  if (y & 1) ans = mult(ans, x, CMOD);
  return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gen(long long x, long long y) {
  if (x == y) return x;
  return x + rng() % (y - x + 1);
}
const int TAM = 1e5 + 100;
const long long INF = LLONG_MAX / 4;
int n;
int a[TAM];
int b[TAM];
string s;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> s;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
  int ans = 0;
  for (int t = 0; t <= TAM; t++) {
    int cont = 0;
    for (int i = 0; i < n; i++) {
      if (t < b[i])
        cont += s[i] - '0';
      else {
        int t2 = t - b[i];
        int p = t2 / a[i];
        if (p % 2 == 0)
          cont += s[i] == '0' ? 1 : 0;
        else
          cont += s[i] - '0';
      }
    }
    ans = max(ans, cont);
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please formulate 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;
const long long MOD = 1e9 + 7;
long long add(long long x, long long y, long long CMOD = MOD) {
  return (x + y + CMOD + CMOD) % CMOD;
}
long long mult(long long x, long long y, long long CMOD = MOD) {
  return add(x, 0) * add(y, 0) % CMOD;
}
long long fast_expo(long long x, long long y, long long CMOD = MOD) {
  if (x == 0) return 0;
  if (y == 0) return 1;
  long long ans = fast_expo(x, y / 2, CMOD);
  ans = mult(ans, ans, CMOD);
  if (y & 1) ans = mult(ans, x, CMOD);
  return ans;
}
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long gen(long long x, long long y) {
  if (x == y) return x;
  return x + rng() % (y - x + 1);
}
const int TAM = 1e5 + 100;
const long long INF = LLONG_MAX / 4;
int n;
int a[TAM];
int b[TAM];
string s;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> s;
  for (int i = 0; i < n; i++) cin >> a[i] >> b[i];
  int ans = 0;
  for (int t = 0; t <= TAM; t++) {
    int cont = 0;
    for (int i = 0; i < n; i++) {
      if (t < b[i])
        cont += s[i] - '0';
      else {
        int t2 = t - b[i];
        int p = t2 / a[i];
        if (p % 2 == 0)
          cont += s[i] == '0' ? 1 : 0;
        else
          cont += s[i] - '0';
      }
    }
    ans = max(ans, cont);
  }
  cout << ans << endl;
  return 0;
}

```