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 int MAXN = 3e5 + 5000;
const long double eps = (1e-12);
const int logn = 21;
const double Pi = acos(-1.0);
const long long pp = 1000000000 + 7;
int lowbit(int x) { return x & (-x); }
const long long INF = 1e9;
long long nums[MAXN];
long long w[MAXN];
long long n, q;
struct node {
  long long sum;
  long long l;
  long long r;
};
node tree[(MAXN << 2) + 5];
inline long long lk(long long p) { return p << 1; }
inline long long rk(long long p) { return (p << 1) | 1; }
long long tag[(MAXN << 2) + 5];
inline void push_up(long long p) {
  tree[p].sum = tree[lk(p)].sum + tree[rk(p)].sum;
}
inline void build(long long p, long long l, long long r) {
  tag[p] = 0;
  tree[p].l = l;
  tree[p].r = r;
  if (l == r) {
    tree[p].sum = w[l];
    return;
  }
  long long mid = (l + r) >> 1;
  build(lk(p), l, mid);
  build(rk(p), mid + 1, r);
  push_up(p);
}
inline void f(long long p, long long l, long long r, long long k) {
  tag[p] = tag[p] + k;
  tree[p].sum = tree[p].sum + k * (r - l + 1);
}
inline void push_down(long long p, long long l, long long r) {
  long long mid = (l + r) >> 1;
  f(lk(p), l, mid, tag[p]);
  f(rk(p), mid + 1, r, tag[p]);
  tag[p] = 0;
}
inline void update(long long nl, long long nr, long long l, long long r,
                   long long p, long long k) {
  if (nl <= l && r <= nr) {
    f(p, l, r, k);
    return;
  }
  push_down(p, l, r);
  long long mid = (l + r) >> 1;
  if (nl <= mid) update(nl, nr, l, mid, lk(p), k);
  if (nr > mid) update(nl, nr, mid + 1, r, rk(p), k);
  push_up(p);
}
inline long long querySum(long long nl, long long nr, long long l, long long r,
                          long long p) {
  if (nr < nl) return 0;
  long long ans = 0;
  if (nl <= l && r <= nr) return tree[p].sum;
  long long mid = (l + r) >> 1;
  push_down(p, l, r);
  if (nl <= mid) ans += querySum(nl, nr, l, mid, lk(p));
  if (nr > mid) ans += querySum(nl, nr, mid + 1, r, rk(p));
  return ans;
}
inline long long queryIdx(long long l, long long r, long long p, long long &tar,
                          long long liml, long long limr) {
  if (liml <= l && r <= limr) {
    if (tree[p].sum < tar) {
      tar -= tree[p].sum;
      return -1;
    }
  }
  if (l == r) {
    if (tree[p].sum >= tar)
      return l;
    else
      tar -= tree[p].sum;
    return -1;
  }
  long long mid = (l + r) >> 1;
  push_down(p, l, r);
  long long ans = -1;
  if (mid < liml) {
    ans = queryIdx(mid + 1, r, rk(p), tar, liml, limr);
  } else if (mid >= limr) {
    ans = queryIdx(l, mid, lk(p), tar, liml, limr);
  } else {
    ans = queryIdx(l, mid, lk(p), tar, liml, limr);
    if (ans == -1) {
      ans = queryIdx(mid + 1, r, rk(p), tar, liml, limr);
    }
  }
  return ans;
}
struct BIT {
  long long S[MAXN] = {}, n = 0;
  void init(long long x) {
    this->n = x;
    for (long long i = 0; i <= n; i++) S[i] = 0;
  }
  void del(long long x) {
    for (long long i = x; i <= n; i += (i & -i)) S[i] = 0;
  }
  void add(long long x, long long v) {
    for (long long i = x; i <= n; i += (i & -i)) {
      S[i] += v;
      S[i] = ((((S[i]) % pp) + pp) % pp);
    }
    return;
  }
  long long ask(long long x) {
    if (x == 0) return 0;
    long long res = 0;
    for (long long i = x; i >= 1; i -= (i & -i)) {
      res += S[i];
      res %= pp;
    }
    return res;
  }
};
BIT bits[3];
signed main() {
  scanf("%lld%lld", &n, &q);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &nums[i]);
  }
  bits[0].init(n);
  bits[1].init(n);
  bits[2].init(n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &w[i]);
    bits[0].add(i, w[i]);
    bits[1].add(i, w[i] * nums[i]);
    bits[2].add(i, w[i] * i);
  }
  build(1, 1, n);
  while (q--) {
    long long x, y;
    scanf("%lld%lld", &x, &y);
    if (x > 0) {
      long long sum = querySum(x, y, 1, n, 1);
      sum = (sum + 1) / 2;
      long long ans = queryIdx(1, n, 1, sum, x, y);
      long long l = x;
      long long anss = 0;
      if (l < ans) {
        long long tmp1 =
            ((((bits[0].ask(ans - 1) - bits[0].ask(l - 1)) % pp) + pp) % pp);
        long long tmp2 =
            ((((bits[1].ask(ans - 1) - bits[1].ask(l - 1)) % pp) + pp) % pp);
        long long tmp3 =
            ((((bits[2].ask(ans - 1) - bits[2].ask(l - 1)) % pp) + pp) % pp);
        anss +=
            ((((((((nums[ans] * tmp1) % pp) + pp) % pp) - tmp2) % pp) + pp) %
             pp);
        anss -= (((((((((ans)*tmp1) % pp) + pp) % pp) - tmp3) % pp) + pp) % pp);
      }
      long long r = y;
      anss = ((((anss) % pp) + pp) % pp);
      if (y > ans) {
        anss +=
            (((((bits[1].ask(r) - bits[1].ask(ans)) -
                ((((nums[ans] *
                    ((((bits[0].ask(r) - bits[0].ask(ans)) % pp) + pp) % pp)) %
                   pp) +
                  pp) %
                 pp)) %
               pp) +
              pp) %
             pp);
        anss -=
            (((((bits[2].ask(r) - bits[2].ask(ans)) -
                (((((ans) *
                    ((((bits[0].ask(r) - bits[0].ask(ans)) % pp) + pp) % pp)) %
                   pp) +
                  pp) %
                 pp)) %
               pp) +
              pp) %
             pp);
      }
      anss = ((((anss) % pp) + pp) % pp);
      printf("%lld\n", anss);
    } else {
      x = -x;
      update(x, x, 1, n, 1, y - w[x]);
      bits[0].add(x, y - w[x]);
      bits[1].add(x, nums[x] * (y - w[x]));
      bits[2].add(x, x * (y - w[x]));
      w[x] = y;
    }
  }
  return 0;
}