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>
#pragma GCC optimize("O3")
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
int getrnd(int l, int r) { return uniform_int_distribution<int>(l, r)(rng); }
template <typename T1, typename T2>
bool relax(T1& a, const T2& b) {
  if (a > b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename T1, typename T2>
bool strain(T1& a, const T2& b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
const int MOD = 1e9 + 7;
int add(int a, int b) {
  a += b;
  if (a >= MOD) a -= MOD;
  if (a < 0) a += MOD;
  return a;
}
int mul(int a, int b) { return 1LL * a * b % MOD; }
const int N = 2e5 + 3;
int n;
ll tree[4 * N], mtree[4 * N];
void build(int v, int tl, int tr, vector<ll>& a, vector<ll>& b) {
  if (tl == tr) {
    tree[v] = a[tl];
    mtree[v] = mul(a[tl], b[tl]);
  } else {
    int tm = (tl + tr) / 2;
    build(2 * v, tl, tm, a, b);
    build(2 * v + 1, tm + 1, tr, a, b);
    tree[v] = tree[2 * v] + tree[2 * v + 1];
    mtree[v] = add(mtree[2 * v], mtree[2 * v + 1]);
  }
}
void update(int v, int tl, int tr, int pos, ll x, vector<ll>& b) {
  if (tl == tr) {
    tree[v] = x;
    mtree[v] = mul(b[tl], x);
  } else {
    int tm = (tl + tr) / 2;
    if (pos <= tm)
      update(2 * v, tl, tm, pos, x, b);
    else
      update(2 * v + 1, tm + 1, tr, pos, x, b);
    tree[v] = tree[2 * v] + tree[2 * v + 1];
    mtree[v] = add(mtree[2 * v], mtree[2 * v + 1]);
  }
}
ll get(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (tl == l && tr == r)
    return tree[v];
  else {
    int tm = (tl + tr) / 2;
    return get(2 * v, tl, tm, l, min(r, tm)) +
           get(2 * v + 1, tm + 1, tr, max(l, tm + 1), r);
  }
}
ll getm(int v, int tl, int tr, int l, int r) {
  if (l > r) return 0;
  if (tl == l && tr == r)
    return mtree[v];
  else {
    int tm = (tl + tr) / 2;
    return add(getm(2 * v, tl, tm, l, min(r, tm)),
               getm(2 * v + 1, tm + 1, tr, max(tm + 1, l), r));
  }
}
int find(int v, int tl, int tr, int l, int r, ll x) {
  if (l > r) return -1;
  if (tl == tr)
    return tl;
  else {
    int tm = (tl + tr) / 2;
    ll s = get(1, 0, n - 1, l, tm);
    if (s >= x)
      return find(2 * v, tl, tm, l, min(r, tm), x);
    else
      return find(2 * v + 1, tm + 1, tr, max(tm + 1, l), r, x - s);
  }
}
void solve() {
  int q;
  cin >> n >> q;
  vector<ll> a(n), w(n), y(n);
  for (int i = 0; i < n; ++i) cin >> a[i], y[i] = add(a[i], -i);
  for (int i = 0; i < n; ++i) cin >> w[i];
  build(1, 0, n - 1, w, y);
  while (q--) {
    int l, r;
    cin >> l >> r;
    if (l < 0) {
      update(1, 0, n - 1, -l - 1, r, y);
    } else {
      --l;
      --r;
      ll s = get(1, 0, n - 1, l, r);
      int pos = find(1, 0, n - 1, l, r, (s + 1) / 2);
      assert(pos != -1);
      ll x = y[pos];
      ll res =
          add(-getm(1, 0, n - 1, l, pos - 1), getm(1, 0, n - 1, pos + 1, r));
      res = add(res, mul(add(get(1, 0, n - 1, l, pos - 1) % MOD,
                             -(get(1, 0, n - 1, pos + 1, r) % MOD)),
                         x));
      cout << res << '\n';
    }
  }
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(nullptr);
  cout.tie(nullptr);
  srand(time(0));
  int t = 1;
  while (t--) solve();
  return 0;
}