#include <bits/stdc++.h>using namespace std;struct Node {Node *ch[2];int s;int flip;int v;int cmp(int k) const {int d = k - ch[0]->s;if(d==1) return -1;     //序列第 k 个找到return d <=0 ? 0 : 1;}void maintain() {s = ch[0]->s + ch[1]->s + 1;}void pushdown() {if(flip) {flip = 0;swap(ch[0],ch[1]);ch[0]->flip = !ch[0]->flip;ch[1]->flip = !ch[1]->flip;}}
};Node *null = new Node();// d = 0 左旋
void rotate(Node* &o,int d) {Node* k = o->ch[d^1];o->ch[d^1] = k->ch[d];k->ch[d] = o;o->maintain();k->maintain();o = k;
}//将序列左数第k个元素选择到根
void splay(Node* &o,int k) {o->pushdown();int d = o->cmp(k);if(d==1) k-=o->ch[0]->s + 1;if(d!=-1) {Node* p = o->ch[d];p->pushdown();int d2 = p->cmp(k);int k2 = (d2==0 ? k : k - p->ch[0]->s - 1);if(d2!=-1) {splay(p->ch[d2],k2);if(d==d2) rotate(o,d^1);else rotate(o->ch[d],d);}rotate(o,d^1);}
}// 合并操作。假定left所有元素小于 right
Node* merge(Node* left,Node* right) {splay(left,left->s);left->ch[1] = right;left->maintain();return left;
}//把 o 前 k 个小结点放到left里面，其他放到ritht里面，如果不够right = null
void split(Node* o,int k,Node* &left,Node* &right) {splay(o,k);left = o;right = o->ch[1];o->ch[1] = null;left->maintain();
}const int maxn = 1e5+10;
struct SplaySequence {int n;Node seq[maxn];Node *root;Node* build(int sz) {if(!sz) return null;Node* L = build(sz/2);Node* o = &seq[++n];o->v = n;o->ch[0] = L;o->ch[1] = build(sz-sz/2-1);o->flip = o ->s = 0;o->maintain();return o;}void init(int sz) {n = 0;null->s = 0;root = build(sz);for(int i = 0; i < sz; i++)printf("%d ",seq[i].v);puts("");}};vector<int> ans;
void print(Node* o) {if(o!=null) {o->pushdown();print(o->ch[0]);ans.push_back(o->v);print(o->ch[1]);}
}void debug(Node* o) {if(o!=null) {o->pushdown();debug(o->ch[0]);printf("%d \n",o->v -1);debug(o->ch[1]);}
}SplaySequence ss;int main()
{int n,m;scanf("%d%d",&n,&m);ss.init(n+1);debug(ss.root);for(int i = 0; i < m; i++) {int a,b;scanf("%d%d",&a,&b);Node* left,*mid,*right,*o;split(ss.root,a,left,o);split(o,b-a+1,mid,right);mid->flip ^=1;ss.root = merge(merge(left,right),mid);}print(ss.root);for(int i = 1; i < (int)ans.size(); i++)printf("%d\n",ans[i]-1);return 0;
}