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;
template <class T>
ostream& prnt(ostream& out, T v) {
  out << v.size() << '\n';
  for (auto e : v) out << e << ' ';
  return out;
}
template <class T>
ostream& operator<<(ostream& out, vector<T> v) {
  return prnt(out, v);
}
template <class T>
ostream& operator<<(ostream& out, set<T> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, map<T1, T2> v) {
  return prnt(out, v);
}
template <class T1, class T2>
ostream& operator<<(ostream& out, pair<T1, T2> p) {
  return out << '(' << p.first << ' ' << p.second << ')';
}
const long long N = 200100;
const long long MARGIN = 1e9;
const long long MOD = 1e9 + 7;
long long n, k, m, a[N], w[N], x, y;
struct first {
  long long val[4 * N];
  first() { memset(val, 0, sizeof val); };
  void upd(long long nod, long long l, long long r, long long pos,
           long long value) {
    if (l == r) {
      val[nod] = value % MOD;
      return;
    }
    long long mid = (l + r) / 2, ans = 0;
    if (pos <= mid)
      upd(2 * nod, l, mid, pos, value);
    else
      upd(2 * nod + 1, mid + 1, r, pos, value);
    val[nod] = val[2 * nod] + val[2 * nod + 1];
    if (val[nod] >= MOD) val[nod] -= MOD;
  }
  long long que(long long nod, long long l, long long r, long long x,
                long long y) {
    if (x <= l && r <= y) return val[nod];
    long long mid = (l + r) / 2, ans = 0;
    if (x <= mid) ans += que(2 * nod, l, mid, x, y), ans %= MOD;
    if (y > mid) ans += que(2 * nod + 1, mid + 1, r, x, y), ans %= MOD;
    return ans;
  }
  void update(long long pos, long long val) { upd(1, 1, n, pos, val); }
  long long query(long long x, long long y) {
    if (x > y) return 0;
    return que(1, 1, n, x, y);
  }
} stl, str;
long long aib[N];
long long lsb(long long x) { return x & (-x); }
long long que(long long pos) {
  long long ret = 0;
  for (; pos; pos -= lsb(pos)) ret += aib[pos];
  return ret;
}
long long query(long long l, long long r) { return que(r) - que(l - 1); }
void update(long long pos, long long val) {
  long long prev = query(pos, pos);
  val = val - prev;
  for (; pos < N; pos += lsb(pos)) aib[pos] += val;
}
long long vv(long long mid, long long l, long long r) {
  return query(l, mid - 1) - query(mid, r);
}
long long cb(long long l, long long r, long long x, long long y) {
  if (l == r) return l;
  long long mid = (l + r + 1) / 2;
  if (vv(mid, x, y) <= 0) return cb(mid, r, x, y);
  return cb(l, mid - 1, x, y);
}
int main() {
  ios_base::sync_with_stdio(false);
  cin >> n >> m;
  for (long long i = 1; i <= n; i++) cin >> a[i];
  for (long long i = 1; i <= n; i++) {
    cin >> w[i];
    stl.update(i, (w[i] * (a[i] - i)) % MOD);
    str.update(i, (w[i] * ((long long)MARGIN - n + i - a[i])) % MOD);
    update(i, w[i]);
  }
  for (long long i = 1; i <= m; i++) {
    cin >> x >> y;
    if (x < 0) {
      x = -x;
      stl.update(x, (y * (a[x] - x)) % MOD);
      str.update(x, (y * ((long long)MARGIN - n + x - a[x])) % MOD);
      update(x, y);
    } else {
      long long l = x, r = y;
      long long fix = cb(l, r, l, r);
      long long ans = 0;
      ans +=
          (str.query(l, fix) -
           (((query(l, fix) % MOD) * ((long long)MARGIN - n + fix - a[fix])) %
            MOD)) %
          MOD;
      ans += (stl.query(fix + 1, r) -
              (((query(fix + 1, r) % MOD) * (a[fix] + 1 - fix - 1)) % MOD)) %
             MOD;
      cout << (ans % MOD + MOD) % MOD << '\n';
    }
  }
}