Problem: There is an infinite line consisting of cells. There are n boxes in some cells of this line. The i-th box stands in the cell a_i and has weight w_i. All a_i are distinct, moreover, a_{i - 1} < a_i holds for all valid i.

You would like to put together some boxes. Putting together boxes with indices in the segment [l, r] means that you will move some of them in such a way that their positions will form some segment [x, x + (r - l)].

In one step you can move any box to a neighboring cell if it isn't occupied by another box (i.e. you can choose i and change a_i by 1, all positions should remain distinct). You spend w_i units of energy moving the box i by one cell. You can move any box any number of times, in arbitrary order.

Sometimes weights of some boxes change, so you have queries of two types: 

  1. id nw — weight w_{id} of the box id becomes nw. 
  2. l r — you should compute the minimum total energy needed to put together boxes with indices in [l, r]. Since the answer can be rather big, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7. Note that the boxes are not moved during the query, you only should compute the answer. 



Note that you should minimize the answer, not its remainder modulo 10^9 + 7. So if you have two possible answers 2 ⋅ 10^9 + 13 and 2 ⋅ 10^9 + 14, you should choose the first one and print 10^9 + 6, even though the remainder of the second answer is 0.

Input

The first line contains two integers n and q (1 ≤ n, q ≤ 2 ⋅ 10^5) — the number of boxes and the number of queries.

The second line contains n integers a_1, a_2, ... a_n (1 ≤ a_i ≤ 10^9) — the positions of the boxes. All a_i are distinct, a_{i - 1} < a_i holds for all valid i.

The third line contains n integers w_1, w_2, ... w_n (1 ≤ w_i ≤ 10^9) — the initial weights of the boxes.

Next q lines describe queries, one query per line.

Each query is described in a single line, containing two integers x and y. If x < 0, then this query is of the first type, where id = -x, nw = y (1 ≤ id ≤ n, 1 ≤ nw ≤ 10^9). If x > 0, then the query is of the second type, where l = x and r = y (1 ≤ l_j ≤ r_j ≤ n). x can not be equal to 0.

Output

For each query of the second type print the answer on a separate line. Since answer can be large, print the remainder it gives when divided by 1000 000 007 = 10^9 + 7.

Example

Input

5 8
1 2 6 7 10
1 1 1 1 2
1 1
1 5
1 3
3 5
-3 5
-1 10
1 4
2 5


Output

0
10
3
4
18
7

Note

Let's go through queries of the example: 

  1. 1\ 1 — there is only one box so we don't need to move anything. 
  2. 1\ 5 — we can move boxes to segment [4, 8]: 1 ⋅ |1 - 4| + 1 ⋅ |2 - 5| + 1 ⋅ |6 - 6| + 1 ⋅ |7 - 7| + 2 ⋅ |10 - 8| = 10. 
  3. 1\ 3 — we can move boxes to segment [1, 3]. 
  4. 3\ 5 — we can move boxes to segment [7, 9]. 
  5. -3\ 5 — w_3 is changed from 1 to 5. 
  6. -1\ 10 — w_1 is changed from 1 to 10. The weights are now equal to w = [10, 1, 5, 1, 2]. 
  7. 1\ 4 — we can move boxes to segment [1, 4]. 
  8. 2\ 5 — we can move boxes to segment [5, 8]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long N = 2e5 + 5;
const long long mod = 1e9 + 7;
long long n, q;
long long a[N], w[N];
template <typename T>
inline void chkmax(T &x, T y) {
  x = max(x, y);
}
template <typename T>
inline void chkmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
inline void read(T &x) {
  T f = 1;
  x = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = -f;
  for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - '0';
  x *= f;
}
inline void inc(long long &x, long long y) {
  x = x + y < mod ? x + y : x + y - mod;
}
inline void dec(long long &x, long long y) {
  x = x - y >= 0 ? x - y : x - y + mod;
}
struct B1 {
  long long c[N];
  inline long long lowbit(long long x) { return x & (-x); }
  inline void mdf(long long x, long long val) {
    for (long long i = x; i <= n; i += lowbit(i)) c[i] += val;
  }
  inline long long qry(long long x) {
    long long ret = 0;
    for (long long i = x; i >= 1; i -= lowbit(i)) ret += c[i];
    return ret;
  }
  inline long long qry(long long l, long long r) {
    if (l > r) return 0;
    return qry(r) - qry(l - 1);
  }
} BitSum;
struct B2 {
  long long c[N];
  inline long long lowbit(long long x) { return x & (-x); }
  inline void mdf(long long x, long long val) {
    val = (val % mod + mod) % mod;
    for (long long i = x; i <= n; i += lowbit(i)) inc(c[i], val);
  }
  inline long long qry(long long x) {
    long long ret = 0;
    for (long long i = x; i >= 1; i -= lowbit(i)) inc(ret, c[i]);
    return ret;
  }
  inline long long qry(long long l, long long r) {
    if (l > r) return 0;
    long long ret = qry(r);
    dec(ret, qry(l - 1));
    return ret;
  }
} BitMul;
inline long long getPos(long long x, long long y) {
  long long ln = x, rn = y, mid, res = 0;
  for (; ln <= rn;) {
    mid = (ln + rn) / 2;
    if (BitSum.qry(x, mid) >= BitSum.qry(mid + 1, y))
      res = mid, rn = mid - 1;
    else
      ln = mid + 1;
  }
  return res;
}
inline void solve(long long x, long long y) {
  if (x == y) {
    puts("0");
    return;
  }
  long long pos = getPos(x, y);
  long long res = 0LL;
  res = (-BitMul.qry(x, pos) +
         (BitSum.qry(x, pos) % mod) * abs(a[pos] - pos) % mod + mod) %
        mod;
  res = (res - (BitSum.qry(pos, y)) % mod * abs(a[pos] - pos) % mod +
         BitMul.qry(pos, y) + mod) %
        mod;
  printf("%lld\n", (res % mod + mod) % mod);
}
signed main() {
  read(n), read(q);
  for (long long i = 1; i <= n; ++i) read(a[i]);
  for (long long i = 1; i <= n; ++i) {
    read(w[i]);
    BitMul.mdf(i, 1LL * w[i] * (a[i] - i));
    BitSum.mdf(i, w[i]);
  }
  while (q--) {
    long long x, y;
    read(x), read(y);
    if (x < 0) {
      x = -x;
      BitSum.mdf(x, -w[x]);
      BitMul.mdf(x, -1LL * w[x] * (a[x] - x));
      w[x] = 1LL * y;
      BitSum.mdf(x, w[x]);
      BitMul.mdf(x, 1LL * w[x] * (a[x] - x));
    } else
      solve(x, y);
  }
  return 0;
}