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;
inline char gc() {
  static char now[1 << 16], *S, *T;
  if (T == S) {
    T = (S = now) + fread(now, 1, 1 << 16, stdin);
    if (T == S) return EOF;
  }
  return *S++;
}
inline int read() {
  int x = 0, f = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = gc();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = gc();
  return x * f;
}
struct node {
  int left, right, lazy;
  long long sum, add;
} tree[110000 << 1];
int num, root, n, m;
inline void build(int &x, int l, int r) {
  x = ++num;
  if (l == r) {
    tree[x].lazy = l;
    return;
  }
  int mid = l + r >> 1;
  build(tree[x].left, l, mid);
  build(tree[x].right, mid + 1, r);
}
inline void update(int x) {
  int l = tree[x].left, r = tree[x].right;
  tree[x].sum = tree[l].sum + tree[r].sum;
  if (tree[l].lazy == tree[r].lazy)
    tree[x].lazy = tree[l].lazy;
  else
    tree[x].lazy = 0;
}
inline void pushdown(int x, int l1, int r1) {
  if (!tree[x].lazy) return;
  int l = tree[x].left, r = tree[x].right, lz = tree[x].lazy;
  tree[l].lazy = lz;
  tree[r].lazy = lz;
  int mid = l1 + r1 >> 1;
  tree[l].add += tree[x].add;
  tree[r].add += tree[x].add;
  tree[l].sum += (long long)(mid - l1 + 1) * tree[x].add;
  tree[r].sum += (long long)(r1 - mid) * tree[x].add;
  tree[x].add = 0;
}
inline void insert1(int x, int l, int r, int l1, int r1, int v) {
  if (l1 <= l && r1 >= r && tree[x].lazy) {
    tree[x].sum += (long long)(r - l + 1) * abs(v - tree[x].lazy);
    tree[x].add += abs(v - tree[x].lazy);
    tree[x].lazy = v;
    return;
  }
  int mid = l + r >> 1;
  pushdown(x, l, r);
  if (l1 <= mid) insert1(tree[x].left, l, mid, l1, r1, v);
  if (r1 > mid) insert1(tree[x].right, mid + 1, r, l1, r1, v);
  update(x);
}
inline long long query(int x, int l, int r, int l1, int r1) {
  if (l1 <= l && r1 >= r) return tree[x].sum;
  int mid = l + r >> 1;
  long long tmp = 0;
  pushdown(x, l, r);
  if (l1 <= mid) tmp += query(tree[x].left, l, mid, l1, r1);
  if (r1 > mid) tmp += query(tree[x].right, mid + 1, r, l1, r1);
  return tmp;
}
inline void print(int x, int l, int r) {
  int mid = l + r >> 1;
  if (tree[x].left) print(tree[x].left, l, mid);
  printf("%d %d %lld %lld %d\n", l, r, tree[x].sum, tree[x].add, tree[x].lazy);
  if (tree[x].right) print(tree[x].right, mid + 1, r);
}
int main() {
  n = read();
  m = read();
  build(root, 1, n);
  for (int i = 1; i <= m; ++i) {
    int op = read(), l = read(), r = read();
    if (op == 1) {
      int v = read();
      insert1(root, 1, n, l, r, v);
    } else
      printf("%lld\n", query(root, 1, n, l, r));
  }
  return 0;
}
