#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:2000000")
#pragma comment(linker, "/HEAP:2000000")
using namespace std;
void print_width(long long x) {
  std::cout << std::fixed;
  std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void printArr(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
void printVector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << '\n';
}
void printVectorPair(std::vector<pair<long long, long long>> v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << '\n';
  ;
  cout << '\n';
}
void initialize(long long arr[], long long n) {
  for (long long i = 0; i <= n; i++) arr[i] = i;
}
long long root(long long arr[], long long i) {
  while (arr[i] != i) {
    arr[i] = arr[arr[i]];
    i = arr[i];
  }
  return i;
}
void Union(long long arr[], long long a, long long b) {
  long long root_a = root(arr, a);
  long long root_b = root(arr, b);
  arr[root_a] = root_b;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power_wm(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
std::vector<long long> vsum(std::vector<long long> a) {
  std::vector<long long> s(a.size());
  s[0] = a[0];
  for (long long i = 1; i <= a.size() - 1; i++) {
    s[i] = s[i - 1] + a[i];
  }
  return s;
}
void time() {}
long long sti(string s) {
  long long ans = 0;
  long long p = 1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    ans = (ans + ((long long)(s[i] - '0') * p) % 1000000007) % 1000000007;
    p = (p * 10) % 1000000007;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  string t;
  cin >> t;
  std::vector<long long> a(n);
  std::vector<long long> b(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cin >> b[i];
  }
  long long ans = 0;
  for (long long i = 0; i <= 10000; i++) {
    long long c = 0;
    for (long long j = 0; j < n; j++) {
      long long time = ((i - b[j]) / a[j]);
      if (i >= b[j] and t[j] == '0' and time % 2 == 0)
        c++;
      else if (i >= b[j] and t[j] == '1' and time % 2 == 1)
        c++;
      if (i < b[j] and t[j] == '1') c++;
    }
    ans = max(ans, c);
  }
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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>
#pragma comment(linker, "/STACK:2000000")
#pragma comment(linker, "/HEAP:2000000")
using namespace std;
void print_width(long long x) {
  std::cout << std::fixed;
  std::cout << std::setprecision(x);
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x = x % p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y = y >> 1;
    x = (x * x) % p;
  }
  return res;
}
void printArr(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cout << a[i] << " ";
  cout << '\n';
}
void printVector(std::vector<long long> v) {
  for (long long i = 0; i < v.size(); i++) cout << v[i] << " ";
  cout << '\n';
}
void printVectorPair(std::vector<pair<long long, long long>> v) {
  for (long long i = 0; i < v.size(); i++)
    cout << v[i].first << " " << v[i].second << '\n';
  ;
  cout << '\n';
}
void initialize(long long arr[], long long n) {
  for (long long i = 0; i <= n; i++) arr[i] = i;
}
long long root(long long arr[], long long i) {
  while (arr[i] != i) {
    arr[i] = arr[arr[i]];
    i = arr[i];
  }
  return i;
}
void Union(long long arr[], long long a, long long b) {
  long long root_a = root(arr, a);
  long long root_b = root(arr, b);
  arr[root_a] = root_b;
}
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long power_wm(long long x, long long y) {
  long long res = 1;
  while (y > 0) {
    if (y & 1) res = (res * x);
    y = y >> 1;
    x = (x * x);
  }
  return res;
}
std::vector<long long> vsum(std::vector<long long> a) {
  std::vector<long long> s(a.size());
  s[0] = a[0];
  for (long long i = 1; i <= a.size() - 1; i++) {
    s[i] = s[i - 1] + a[i];
  }
  return s;
}
void time() {}
long long sti(string s) {
  long long ans = 0;
  long long p = 1;
  for (long long i = s.size() - 1; i >= 0; i--) {
    ans = (ans + ((long long)(s[i] - '0') * p) % 1000000007) % 1000000007;
    p = (p * 10) % 1000000007;
  }
  return ans;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long n;
  cin >> n;
  string t;
  cin >> t;
  std::vector<long long> a(n);
  std::vector<long long> b(n);
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
    cin >> b[i];
  }
  long long ans = 0;
  for (long long i = 0; i <= 10000; i++) {
    long long c = 0;
    for (long long j = 0; j < n; j++) {
      long long time = ((i - b[j]) / a[j]);
      if (i >= b[j] and t[j] == '0' and time % 2 == 0)
        c++;
      else if (i >= b[j] and t[j] == '1' and time % 2 == 1)
        c++;
      if (i < b[j] and t[j] == '1') c++;
    }
    ans = max(ans, c);
  }
  cout << ans;
  return 0;
}

```