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 INF = 0x3f3f3f3f;
const long long INFL = 0x3f3f3f3f3f3f3f3fLL;
const long double eps = 1e-8;
const int MAXN = 100000 + 5;
pair<long double, long double> X[MAXN << 2];
inline pair<long double, long double> calc(const int& a, const int& b) {
  long double x = (long double)(b - a) / (long double)a;
  return make_pair(x, x);
}
inline void pushUp(const int& rt) {
  X[rt].first = X[rt << 1].first * X[rt << 1 | 1].first;
  X[rt].second = X[rt << 1].second + X[rt << 1 | 1].second * X[rt << 1].first;
}
void build(int l, int r, int rt) {
  static int ax, bx;
  if (l == r) {
    scanf("%d %d", &ax, &bx);
    X[rt] = calc(ax, bx);
    return;
  }
  int mid = (l + r) >> 1;
  build(l, mid, rt << 1);
  build(mid + 1, r, rt << 1 | 1);
  pushUp(rt);
}
void update(const int& p, const pair<long double, long double>& v, int l, int r,
            int rt) {
  if (l == r) {
    X[rt] = v;
    return;
  }
  int mid = (l + r) >> 1;
  if (p <= mid)
    update(p, v, l, mid, rt << 1);
  else
    update(p, v, mid + 1, r, rt << 1 | 1);
  pushUp(rt);
}
pair<long double, long double> query(const int& L, const int& R, int l, int r,
                                     int rt) {
  static pair<long double, long double> z =
      make_pair((long double)1.0, (long double)0.0);
  if (L <= l && r <= R) return X[rt];
  int mid = (l + r) >> 1;
  pair<long double, long double> x = z, y = z, ret;
  if (L <= mid) x = query(L, R, l, mid, rt << 1);
  if (R > mid) y = query(L, R, mid + 1, r, rt << 1 | 1);
  ret.first = x.first * y.first;
  ret.second = x.second + y.second * x.first;
  return ret;
}
int main() {
  static int n, q, op, x, ax, bx, lx, rx;
  while (~scanf("%d %d", &n, &q)) {
    build(1, n, 1);
    while (q--) {
      scanf("%d", &op);
      if (op == 1) {
        scanf("%d %d %d", &x, &ax, &bx);
        update(x, calc(ax, bx), 1, n, 1);
      } else {
        scanf("%d %d", &lx, &rx);
        long double ans = query(lx, rx, 1, n, 1).second;
        if (ans != ans || ans >= 1e20)
          ans = 0.0;
        else
          ans = 1 / (1 + ans);
        cout << fixed << setprecision(8) << ans << "\n";
      }
    }
  }
  return 0;
}