New Delhi, Oct 10, 2019-
Chinese President Xi Jinping will spend nearly 24 hours in India during his second informal bilateral with Prime Minister Narendra Modi at Mahabalipuram in Tamil Nadu.
According to the itinerary, Xi will arrive at the Chennai International Airport at 2.10 p.m. on Friday.
He will depart for Mahabalipuram, which is 50 kms away from Chennai, at 4.00 p.m.
At 5.00 p.m., he will be given a guided tour of three monuments — Arjuna’s Penance, Panch Rathas and the Shore Temple — at Mahabalipuram
At 6.00 p.m., he will witness a cultural performance by the Kalakshetra team at the Shore Temple.
At 6.45 p.m., he will attend a dinner hosted by Prime Minister Narendra Modi.
On Saturday, the second day of the visit, Xi will arrive at the Taj Fisherman’s Cove Resort & Spa at 9.50 a.m.
At 10.00 a.m., the two leaders will engage in a tete-e-tete at the Machan, Taj Fisherman’s Cove Resort & Spa.
At 10.50 a.m., the two sides will hold delegation-level talks at the Tango Hall, Taj Fisherman’s Cove Resort & Spa.
At 11.45 a.m., Xi will attend a lunch hosted by Prime Minister Modi.
At 12.45 p.m., he will depart for the Chennai International Airport.
At 13.30 p.m., he will emplane for his next destination.
The media will have access to the Media Centre at Sheraton Grand, Chennai, according to the itinerary shared by the Ministry of External Affairs (MEA). (Agency)