DZY loves colors, and he enjoys painting.

On a colorful day, DZY gets a colorful ribbon, which consists of n units (they are numbered from 1 to n from left to right). The color of the i-th unit of the ribbon is i at first. It is colorful enough, but we still consider that the colorfulness of each unit is 0 at first.

DZY loves painting, we know. He takes up a paintbrush with color x and uses it to draw a line on the ribbon. In such a case some contiguous units are painted. Imagine that the color of unit i currently is y. When it is painted by this paintbrush, the color of the unit becomes x, and the colorfulness of the unit increases by |x - y|.

DZY wants to perform m operations, each operation can be one of the following:

  1. Paint all the units with numbers between l and r (both inclusive) with color x. 
  2. Ask the sum of colorfulness of the units between l and r (both inclusive). 



Can you help DZY?

Input

The first line contains two space-separated integers n, m (1 ≤ n, m ≤ 105).

Each of the next m lines begins with a integer type (1 ≤ type ≤ 2), which represents the type of this operation.

If type = 1, there will be 3 more integers l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 108) in this line, describing an operation 1.

If type = 2, there will be 2 more integers l, r (1 ≤ l ≤ r ≤ n) in this line, describing an operation 2.

Output

For each operation 2, print a line containing the answer — sum of colorfulness.

Examples

Input

3 3
1 1 2 4
1 2 3 5
2 1 3


Output

8


Input

3 4
1 1 3 4
2 1 1
2 2 2
2 3 3


Output

3
2
1


Input

10 6
1 1 5 3
1 2 7 9
1 10 10 11
1 3 8 12
1 1 10 3
2 1 10


Output

129

Note

In the first sample, the color of each unit is initially [1, 2, 3], and the colorfulness is [0, 0, 0].

After the first operation, colors become [4, 4, 3], colorfulness become [3, 2, 0].

After the second operation, colors become [4, 5, 5], colorfulness become [3, 3, 2].

So the answer to the only operation of type 2 is 8.
#include <bits/stdc++.h>
using namespace std;
const int N = 1e5 + 5;
int n, m;
struct node {
  long long sum;
  int x;
  bool f;
};
node tree[N * 4], lazy[N * 4];
inline void build(int st = 0, int en = n - 1, int root = 1) {
  if (st == en) {
    tree[root].f = 1;
    tree[root].x = st + 1;
    return;
  }
  build(st, ((st + en) >> 1), root << 1);
  build(((st + en) >> 1) + 1, en, (root << 1) | 1);
}
inline void apply(int root, int st, int en) {
  if (lazy[root].x == 0) return;
  int L = root << 1, R = (root << 1) | 1;
  tree[L].f = tree[R].f = 1;
  lazy[R].x = lazy[L].x = tree[L].x = tree[R].x = lazy[root].x;
  lazy[R].sum += lazy[root].sum;
  lazy[L].sum += lazy[root].sum;
  tree[L].sum += (((st + en) >> 1) - st + 1) * 1LL * lazy[root].sum;
  tree[R].sum += (en - ((st + en) >> 1)) * 1LL * lazy[root].sum;
  lazy[root].sum = 0;
}
void update(int L, int R, int x, int st = 0, int en = n - 1, int root = 1) {
  if (L > en || R < st) return;
  if (st >= L && en <= R && tree[root].f) {
    tree[root].sum += (en - st + 1) * 1LL * abs(x - tree[root].x);
    lazy[root].sum += abs(x - tree[root].x);
    tree[root].x = x;
    lazy[root].x = x;
    return;
  }
  if (tree[root].f) apply(root, st, en);
  update(L, R, x, st, ((st + en) >> 1), root << 1);
  update(L, R, x, ((st + en) >> 1) + 1, en, (root << 1) | 1);
  if (tree[root << 1].f && tree[(root << 1) | 1].f &&
      tree[root << 1].x == tree[(root << 1) | 1].x) {
    tree[root].f = 1;
    tree[root].x = tree[root << 1].x;
  } else {
    tree[root].f = 0;
    lazy[root].x = 0;
  }
  tree[root].sum = tree[(root << 1)].sum + tree[(root << 1) | 1].sum;
}
long long query(int L, int R, int st = 0, int en = n - 1, int root = 1) {
  if (L > en || R < st) return 0;
  if (st >= L && en <= R) return tree[root].sum;
  if (tree[root].f) apply(root, st, en);
  return query(L, R, st, ((st + en) >> 1), root << 1) +
         query(L, R, ((st + en) >> 1) + 1, en, (root << 1) | 1);
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  ;
  cin >> n >> m;
  build();
  while (m--) {
    int t, L, R;
    cin >> t >> L >> R;
    L--, R--;
    if (t == 1) {
      int x;
      cin >> x;
      update(L, R, x);
    } else {
      long long ret = query(L, R);
      cout << ret << '\n';
    }
  }
  return 0;
}
