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 MOD = (int)1e9 + 7;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
const int maxn = 100000 + 10;
int n, q;
int a[maxn];
int b[maxn];
struct Node {
  double x, y;
  Node() : x(-1), y(-1) {}
  Node(double x, double y) : x(x), y(y) {}
  Node(const Node& rhs) : x(rhs.x), y(rhs.y) {}
} nil;
Node merge(Node a, Node b) {
  if (a.x == nil.x) return b;
  if (b.x == nil.x) return a;
  Node res;
  res.x = a.x * b.x / (1.0 - a.y + a.y * b.x);
  res.y = b.y + (1.0 - b.y) * a.y * b.x / (1.0 - a.y + a.y * b.x);
  return res;
}
Node st[maxn << 2];
void build(int p, int L, int R) {
  if (L == R) {
    st[p].x = st[p].y = 1.0 * a[L] / b[L];
    return;
  }
  build(p << 1, L, L + R >> 1);
  build(p << 1 | 1, (L + R >> 1) + 1, R);
  st[p] = merge(st[p << 1], st[p << 1 | 1]);
}
void upd(int p, int i, int L, int R) {
  if (i < L || i > R) return;
  if (L == R) {
    st[p].x = st[p].y = 1.0 * a[L] / b[L];
    return;
  }
  upd(p << 1, i, L, L + R >> 1);
  upd(p << 1 | 1, i, (L + R >> 1) + 1, R);
  st[p] = merge(st[p << 1], st[p << 1 | 1]);
}
Node query(int p, int l, int r, int L, int R) {
  if (l > R || r < L) return nil;
  if (l <= L && r >= R) return st[p];
  return merge(query(p << 1, l, r, L, L + R >> 1),
               query(p << 1 | 1, l, r, (L + R >> 1) + 1, R));
}
void solve() {
  cin >> n >> q;
  for (int i = (0); i < (n); i++) {
    cin >> a[i] >> b[i];
  }
  build(1, 0, n - 1);
  while (q--) {
    int op;
    cin >> op;
    if (op == 1) {
      int ix;
      cin >> ix;
      ix--;
      cin >> a[ix] >> b[ix];
      upd(1, ix, 0, n - 1);
    } else {
      int l, r;
      cin >> l >> r;
      l--;
      r--;
      Node res = query(1, l, r, 0, n - 1);
      cout << fixed << setprecision(9) << res.x << "\n";
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  solve();
  return 0;
}