Problem: There are n casinos lined in a row. If Memory plays at casino i, he has probability pi to win and move to the casino on the right (i + 1) or exit the row (if i = n), and a probability 1 - pi to lose and move to the casino on the left (i - 1) or also exit the row (if i = 1). 

We say that Memory dominates on the interval i... j if he completes a walk such that,

  * He starts on casino i. 
  * He never looses in casino i. 
  * He finishes his walk by winning in casino j. 



Note that Memory can still walk left of the 1-st casino and right of the casino n and that always finishes the process.

Now Memory has some requests, in one of the following forms:

  * 1 i a b: Set <image>. 
  * 2 l r: Print the probability that Memory will dominate on the interval l... r, i.e. compute the probability that Memory will first leave the segment l... r after winning at casino r, if she starts in casino l. 



It is guaranteed that at any moment of time p is a non-decreasing sequence, i.e. pi ≤ pi + 1 for all i from 1 to n - 1.

Please help Memory by answering all his requests!

Input

The first line of the input contains two integers n and q(1 ≤ n, q ≤ 100 000), — number of casinos and number of requests respectively.

The next n lines each contain integers ai and bi (1 ≤ ai < bi ≤ 109) — <image> is the probability pi of winning in casino i.

The next q lines each contain queries of one of the types specified above (1 ≤ a < b ≤ 109, 1 ≤ i ≤ n, 1 ≤ l ≤ r ≤ n).

It's guaranteed that there will be at least one query of type 2, i.e. the output will be non-empty. Additionally, it is guaranteed that p forms a non-decreasing sequence at all times.

Output

Print a real number for every request of type 2 — the probability that boy will "dominate" on that interval. Your answer will be considered correct if its absolute error does not exceed 10 - 4.

Namely: let's assume that one of your answers is a, and the corresponding answer of the jury is b. The checker program will consider your answer correct if |a - b| ≤ 10 - 4.

Example

Input

3 13
1 3
1 2
2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3
1 2 2 3
2 1 1
2 1 2
2 1 3
2 2 2
2 2 3
2 3 3


Output

0.3333333333
0.2000000000
0.1666666667
0.5000000000
0.4000000000
0.6666666667
0.3333333333
0.2500000000
0.2222222222
0.6666666667
0.5714285714
0.6666666667

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 120000;
struct sad {
  double m[2][2];
  void _set(double a, double b, double c, double d) {
    m[0][0] = a;
    m[0][1] = b;
    m[1][0] = c;
    m[1][1] = d;
  }
  sad operator*(const sad &b) const {
    sad c;
    c._set(0, 0, 0, 0);
    for (int i = 0; i < 2; i++)
      for (int j = 0; j < 2; j++)
        for (int k = 0; k < 2; k++) {
          if (m[i][k] > (1e18) || b.m[k][j] > (1e18) || c.m[i][j] > (1e18))
            c.m[i][j] = (1e18) + 1;
          else
            c.m[i][j] += m[i][k] * b.m[k][j];
        }
    return c;
  }
} rs[maxn << 2];
int a[maxn], b[maxn];
void pushup(int rt) { rs[rt] = rs[rt << 1 | 1] * rs[rt << 1]; }
void build(int l, int r, int rt) {
  if (l == r) {
    rs[rt]._set(1.0 * b[l] / a[l], 1 - 1.0 * b[l] / a[l], 1, 0);
    return;
  }
  int m = l + r >> 1;
  build(l, m, rt << 1);
  build(m + 1, r, rt << 1 | 1);
  pushup(rt);
}
void update(int p, int l, int r, int rt) {
  if (l == r) {
    rs[rt]._set(1.0 * b[l] / a[l], 1 - 1.0 * b[l] / a[l], 1, 0);
    return;
  }
  int m = l + r >> 1;
  if (p <= m)
    update(p, l, m, rt << 1);
  else
    update(p, m + 1, r, rt << 1 | 1);
  pushup(rt);
}
sad ans;
void query(int L, int R, int l, int r, int rt) {
  if (L <= l & r <= R) {
    ans = rs[rt] * ans;
    return;
  }
  int m = l + r >> 1;
  if (L <= m) query(L, R, l, m, rt << 1);
  if (m < R) query(L, R, m + 1, r, rt << 1 | 1);
}
int main() {
  int n, q, p, l, r;
  while (~scanf("%d%d", &n, &q)) {
    for (int i = 1; i <= n; i++) scanf("%d%d", &a[i], &b[i]);
    build(1, n, 1);
    while (q--) {
      scanf("%d%d", &p, &l);
      if (p == 1) {
        scanf("%d%d", &a[l], &b[l]);
        update(l, 1, n, 1);
      } else {
        scanf("%d", &r);
        ans._set(1, 0, 0, 1);
        query(l, r, 1, n, 1);
        printf("%f\n", ans.m[0][0] > (1e18) ? 0 : 1.0 / ans.m[0][0]);
      }
    }
  }
  return 0;
}