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;
#pragma GCC target("avx2")
#pragma GCC optimization("O3")
#pragma GCC optimization("unroll-loops")
struct greateri {
  template <class T>
  bool operator()(T const &a, T const &b) const {
    return a > b;
  }
};
bool ka;
void setIO(string s) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  freopen((s + ".in").c_str(), "r", stdin);
  freopen((s + ".out").c_str(), "w", stdout);
}
long long seg[1800001], seg1[1800001];
long long seg2[1800001], seg3[1800001];
long long cura[200001];
vector<long long> initial, fat, nowi, nowi1;
long long n;
long long BIT[1800001];
long long curi[1800001], casi[1800001];
void build(long long p, long long l, long long r, long long yo) {
  if (l == r) {
    if (!yo) {
      if (!l)
        seg[p] = 0;
      else
        seg[p] = (nowi[l] *
                  ((fat[l] - (fat[l - 1] + 1) + 1000000007) % 1000000007)) %
                 1000000007;
    } else {
      if (l == (n - 1))
        seg1[p] = 0;
      else
        seg1[p] = (nowi1[l] *
                   ((fat[l + 1] - (fat[l] + 1) + 1000000007) % 1000000007)) %
                  1000000007;
    }
    return;
  }
  long long med = (l + r) / 2;
  build(p * 2, l, med, yo);
  build(p * 2 + 1, med + 1, r, yo);
  if (!yo) {
    seg[p] = (seg[p * 2] + seg[p * 2 + 1]) % 1000000007;
  } else
    seg1[p] = (seg1[p * 2] + seg1[p * 2 + 1]) % 1000000007;
}
void build3(long long p, long long l, long long r) {
  if (l == r) {
    if (!l)
      seg2[p] = 0;
    else
      seg2[p] = (fat[l] - (fat[l - 1] + 1) + 1000000007) % 1000000007;
    if (l == (n - 1))
      seg3[p] = 0;
    else
      seg3[p] = (fat[l + 1] - (fat[l] + 1) + 1000000007) % 1000000007;
    cura[l] = seg2[p];
    casi[l] = seg3[p];
    return;
  }
  long long med = (l + r) / 2;
  build3(p * 2, l, med);
  build3(p * 2 + 1, med + 1, r);
  seg2[p] = (seg2[p * 2] + seg2[p * 2 + 1]);
  seg3[p] = (seg3[p * 2] + seg3[p * 2 + 1]);
}
pair<long long, long long> queri(long long p, long long l, long long r,
                                 long long l1, long long r1) {
  if (l > r1 || l1 > r) return {0, 0};
  if (l >= l1 && r <= r1) return {seg2[p], seg3[p]};
  long long med = (l + r) / 2;
  auto u = queri(p * 2, l, med, l1, r1);
  auto v = queri(p * 2 + 1, med + 1, r, l1, r1);
  return {(u.first + v.first), (u.second + v.second)};
}
vector<long long> bos, fos;
long long lazy[1800001], lazy1[1800001];
long long l;
void prop(long long node, long long l, long long r, long long in) {
  if (l != r) {
    long long med = (l + r) / 2;
    auto u = 0;
    auto v = 0;
    if (!in) {
      if (!l)
        u = fos[med];
      else
        u = (fos[med] - fos[l - 1] + 1000000007) % 1000000007;
      if (!(med + 1))
        v = fos[r];
      else
        v = (fos[r] - fos[med] + 1000000007) % 1000000007;
      long long soli = (((u)*lazy[node]) % 1000000007);
      long long koli = (((v)*lazy[node]) % 1000000007);
      seg[node * 2] = (seg[node * 2] + soli + 1000000007) % 1000000007;
      seg[node * 2 + 1] = (seg[node * 2 + 1] + koli + 1000000007) % 1000000007;
      lazy[node * 2] = (lazy[node * 2] + lazy[node]) % 1000000007;
      lazy[node * 2 + 1] = (lazy[node] + lazy[node * 2 + 1]) % 1000000007;
      lazy[node] = 0;
    } else {
      if (!l)
        u = bos[med];
      else
        u = (bos[med] - bos[l - 1] + 1000000007) % 1000000007;
      if (!(med + 1))
        v = bos[r];
      else
        v = (bos[r] - bos[med] + 1000000007) % 1000000007;
      long long soli = (((u)*lazy1[node]) % 1000000007);
      long long koli = (((v)*lazy1[node]) % 1000000007);
      seg1[node * 2] = (seg1[node * 2] + soli + 1000000007) % 1000000007;
      seg1[node * 2 + 1] =
          (seg1[node * 2 + 1] + koli + 1000000007) % 1000000007;
      lazy1[node * 2] = (lazy1[node * 2] + lazy1[node]) % 1000000007;
      lazy1[node * 2 + 1] = (lazy1[node * 2 + 1] + lazy1[node]) % 1000000007;
      lazy1[node] = 0;
    }
  }
}
void badel(long long p, long long l, long long r, long long l1, long long r1,
           long long val, long long sasy) {
  if (l1 > r1) return;
  if (l > r1 || r < l1) return;
  if (l >= l1 && r <= r1) {
    auto cur = 0;
    if (!sasy) {
      if (!l)
        cur = fos[r];
      else
        cur = (fos[r] - fos[l - 1] + 1000000007) % 1000000007;
      long long dal = (cur * val) % 1000000007;
      seg[p] = (seg[p] + dal + 1000000007) % 1000000007;
      lazy[p] = (lazy[p] + val) % 1000000007;
    } else {
      if (!l)
        cur = bos[r];
      else
        cur = (bos[r] - bos[l - 1] + 1000000007) % 1000000007;
      long long dal = (cur * val) % 1000000007;
      seg1[p] = (seg1[p] + dal + 1000000007) % 1000000007;
      lazy1[p] = (lazy1[p] + val) % 1000000007;
    }
    prop(p, l, r, sasy);
    return;
  }
  prop(p, l, r, sasy);
  long long med = (l + r) / 2;
  badel(p * 2, l, med, l1, r1, val, sasy);
  badel(p * 2 + 1, med + 1, r, l1, r1, val, sasy);
  if (!sasy) {
    seg[p] = (seg[p * 2] + seg[p * 2 + 1] + 1000000007) % 1000000007;
  } else
    seg1[p] = (seg1[p * 2] + seg1[p * 2 + 1] + 1000000007) % 1000000007;
}
long long lawej(long long p, long long l, long long r, long long l1,
                long long r1, long long sasy) {
  if (l1 > r1) return 0;
  if (l > r1 || r < l1) return 0;
  if (l >= l1 && r <= r1) {
    if (!sasy) return seg[p] % 1000000007;
    return seg1[p] % 1000000007;
  }
  prop(p, l, r, sasy);
  long long med = (l + r) / 2;
  return (lawej(p * 2, l, med, l1, r1, sasy) +
          lawej(p * 2 + 1, med + 1, r, l1, r1, sasy) + 1000000007) %
         1000000007;
}
void update(long long index, long long val) {
  for (index++; index <= n; index += (index & (-index))) {
    BIT[index] += val;
  }
}
long long query(long long index) {
  long long result = 0;
  for (index++; index > 0; index -= (index & (-index))) {
    result = (result + BIT[index]);
  }
  return result;
}
long long range(long long l, long long r) {
  if (l > r) return 0;
  return ((query(r) - query(l - 1)));
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long q;
  cin >> n >> q;
  if (n == 200000) ka = true;
  l = log2(n);
  int index = 0;
  for (long long i = 0; i < n; ++i) {
    long long xo;
    cin >> xo;
    fat.push_back(xo);
  }
  for (long long i = 0; i < n; ++i) {
    long long yo;
    cin >> yo;
    initial.push_back(yo);
    curi[i] = yo;
    update(i, yo);
  }
  long long cumi = 0;
  vector<long long> useful;
  for (long long i = 0; i < n; ++i) {
    nowi.push_back(cumi);
    cumi = (cumi + initial[i]) % 1000000007;
    useful.push_back(cumi);
  }
  cumi = 0;
  for (long long i = n - 1; i >= 0; --i) {
    nowi1.push_back(cumi);
    cumi = (cumi + initial[i]) % 1000000007;
  }
  reverse(nowi1.begin(), nowi1.end());
  build(1, 0, n - 1, 0);
  build(1, 0, n - 1, 1);
  build3(1, 0, n - 1);
  cumi = 0;
  for (int i = 0; i < n; ++i) {
    cumi = (cumi + cura[i]) % 1000000007;
    fos.push_back(cumi);
  }
  cumi = 0;
  for (int i = 0; i < n; ++i) {
    cumi = (cumi + casi[i]) % 1000000007;
    bos.push_back(cumi);
  }
  while (q--) {
    long long x;
    cin >> x;
    index++;
    if (x < 0) {
      x = -x - 1;
      long long newi;
      long long lasti = curi[x];
      cin >> newi;
      update(x, newi - lasti);
      badel(1, 0, n - 1, x + 1, n - 1, newi - lasti, 0);
      badel(1, 0, n - 1, 0, x - 1, newi - lasti, 1);
      curi[x] = newi;
    } else {
      long long r;
      cin >> r;
      long long l = x;
      l--, r--;
      long long low = l;
      long long high = r;
      long long besti = l;
      while (low <= high) {
        long long med = (low + high) / 2;
        long long k = range(l, med - 1);
        long long f = range(med, r);
        if ((k - f) < 0) {
          besti = med;
          low = med + 1;
        } else {
          high = med - 1;
        }
      }
      long long fol = 0;
      if (!l)
        fol = fos[besti] % 1000000007;
      else
        fol = (fos[besti] - fos[l - 1] + 1000000007) % 1000000007;
      long long sol = 0;
      if (!besti)
        sol = bos[r] % 1000000007;
      else
        sol = (bos[r] - bos[besti - 1] + 1000000007) % 1000000007;
      auto kal = range(0, l - 1) % 1000000007;
      auto nal = range(r + 1, n - 1) % 1000000007;
      auto fin1 = lawej(1, 0, n - 1, l, besti, 0) % 1000000007;
      auto fin2 = lawej(1, 0, n - 1, besti, r, 1) % 1000000007;
      fin1 = ((fin1 - ((fol * kal) % 1000000007)) + 1000000007) % 1000000007;
      fin2 = ((fin2 - ((sol * nal) % 1000000007)) + 1000000007) % 1000000007;
      cout << (fin1 + fin2) % 1000000007 << '\n';
    }
  }
  return 0;
}