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;
struct Node {
  int l, r, col;
  bool same;
  long long sum, add;
} tree[100010 << 2];
void color(long long add, int x, int l, int r, int rt) {
  tree[rt].add += add;
  tree[rt].same = true;
  tree[rt].sum += (long long)(r - l + 1) * (long long)add;
  tree[rt].col = x;
}
void pushDown(int l, int r, int rt) {
  int m = (l + r) / 2;
  if (tree[rt].add) {
    color(tree[rt].add, tree[rt].col, l, m, rt << 1);
    color(tree[rt].add, tree[rt].col, m + 1, r, rt << 1 | 1);
    tree[rt].add = 0;
  }
}
void pushUp(int rt) {
  tree[rt].sum = tree[rt << 1].sum + tree[rt << 1 | 1].sum;
  if (tree[rt << 1].col == tree[rt << 1 | 1].col &&
      (tree[rt << 1].same && tree[rt << 1 | 1].same)) {
    tree[rt].same = true;
    tree[rt].col = tree[rt << 1].col;
  } else
    tree[rt].same = false;
}
void buildTree(int l, int r, int rt) {
  if (l == r) {
    tree[rt].col = l;
    tree[rt].same = true;
    tree[rt].sum = 0;
    return;
  }
  int m = (l + r) / 2;
  buildTree(l, m, rt << 1);
  buildTree(m + 1, r, rt << 1 | 1);
  pushUp(rt);
}
void update(int L, int R, int x, int l, int r, int rt) {
  if (L == l && R == r && tree[rt].same) {
    color((long long)abs((long long)x - (long long)tree[rt].col), x, l, r, rt);
    return;
  }
  pushDown(l, r, rt);
  int m = (l + r) / 2;
  if (R <= m)
    update(L, R, x, l, m, rt << 1);
  else if (L > m)
    update(L, R, x, m + 1, r, rt << 1 | 1);
  else {
    update(L, m, x, l, m, rt << 1);
    update(m + 1, R, x, m + 1, r, rt << 1 | 1);
  }
  pushUp(rt);
}
long long query(int L, int R, int l, int r, int rt) {
  if (l == L && r == R) {
    return tree[rt].sum;
  }
  pushDown(l, r, rt);
  int m = (l + r) / 2;
  long long res = 0;
  if (R <= m)
    res += query(L, R, l, m, rt << 1);
  else if (L > m)
    res += query(L, R, m + 1, r, rt << 1 | 1);
  else {
    res += query(L, m, l, m, rt << 1);
    res += query(m + 1, R, m + 1, r, rt << 1 | 1);
  }
  return res;
}
int main() {
  int n, m, str, a, b, c;
  scanf("%d%d", &n, &m);
  buildTree(1, n, 1);
  while (m--) {
    scanf("%d", &str);
    if (str == 1) {
      scanf("%d%d%d", &a, &b, &c);
      update(a, b, c, 1, n, 1);
    } else if (str == 2) {
      scanf("%d%d", &a, &b);
      long long res = query(a, b, 1, n, 1);
      printf("%I64d\n", res);
    }
  }
  return 0;
}
